# Problem R*5.1 - Equivalence of twos forms of 2nd exactness condition

From Mtg 22-6

## Given

 {\displaystyle \displaystyle {\begin{aligned}g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0\end{aligned}}} (1.1)
 {\displaystyle \displaystyle {\begin{aligned}g_{0}={\frac {\partial G}{\partial y}}\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}g_{1}={\frac {\partial G}{\partial y'}}\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}g_{2}={\frac {\partial ^{2}G}{\partial y''}}\end{aligned}}}

## Find

Equivalence of the form of 2nd exactness condition:

 {\displaystyle \displaystyle {\begin{aligned}f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y}+(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0\end{aligned}}}

## Solution

- Solved on our own


Let's assume that p=y' and q=y.

 {\displaystyle \displaystyle {\begin{aligned}g_{0}&={\frac {\partial G}{\partial y}}\\&={\frac {\partial (g+fy'')}{\partial y}}&=g_{y}+f_{y}q\end{aligned}}} (1.1)
 {\displaystyle \displaystyle {\begin{aligned}g_{1}&={\frac {\partial G}{\partial y'}}\\&={\frac {\partial (g+fy'')}{\partial p}}&=g_{p}+f_{p}q\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}{\frac {dg_{1}}{dx}}&={\frac {d(g_{p}+f_{p}q)}{dx}}\\&={\frac {\partial (g_{p}+f_{p}q)}{\partial x}}+{\frac {\partial (g_{p}+f_{p}q)}{\partial y}}{\frac {dy}{dx}}+{\frac {\partial (g_{p}+f_{p}q)}{\partial p}}{\frac {dp}{dx}}\\&=g_{px}+f_{px}q+(g_{py}+f_{yy}q)p+(g_{pp}+f_{yp}q)q\\&=g_{px}+g_{py}p+g_{pp}q+f_{px}q+f_{yy}qp+f_{yp}q^{2}\end{aligned}}} (1.2)
 {\displaystyle \displaystyle {\begin{aligned}g_{2}&={\frac {\partial G}{\partial y''}}\\&={\frac {\partial (g+fy'')}{\partial q}}&=f\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}{\frac {dg_{2}}{dx}}&={\frac {d}{dx}}({\frac {d(f)}{dx}})\\&={\frac {d}{dx}}(f_{x}+f_{y}p+f_{p}q)\\&={\frac {\partial (f_{x}+f_{y}p+f_{p}q)}{\partial x}}+{\frac {\partial (f_{x}+f_{y}p+f_{p}q)}{\partial y}}{\frac {dy}{dx}}+{\frac {\partial (f_{x}+f_{y}p+f_{p}q)}{\partial p}}{\frac {dp}{dx}}\\&=f_{xx}+f_{xy}p+f_{xp}q+f_{yx}p+f_{yy}p^{2}+f_{yp}pq+f_{y}q+f_{px}q+f_{py}pq+f_{pp}q^{2}\\&=f_{xx}+2f_{xy}p+f_{yy}p^{2}+2f_{xp}q+2f_{yp}pq+f_{pp}q^{2}\end{aligned}}} (1.3)

As a result,

 {\displaystyle \displaystyle {\begin{aligned}g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}&=g_{y}+f_{y}q-(g_{px}+g_{py}p+g_{pp}q+f_{px}q+f_{yy}qp+f_{yp}q^{2})+f_{xx}+2f_{xy}p+f_{yy}p^{2}+2f_{xp}q+2f_{yp}pq+f_{pp}q^{2}\\&=f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y}+(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0\end{aligned}}}

# Problem R*5.2 - Verification of the exactness of Legendre and Hermite equations

From Mtg 27-1

## Given

Legendre equation:

 ${\displaystyle \displaystyle G=(1-x^{2})y^{''}-2xy^{\prime }+n(n+1)y=0}$ (2.1)

Hermite equation:

 ${\displaystyle \displaystyle y^{''}-2xy^{\prime }+2ny=0}$ (2.2)

## Find

1. Verify the exactness of the designated L2-ODE-VC (Eq. 2.1, 2.2).
2. If Hermite equation is not exact, check whether it is in power form, and see whether it can be made exact using IFM with ${\displaystyle \displaystyle h(x,y)=x^{m}y^{n}}$
3. The first few Hermite polynomials are given as below.

 {\displaystyle \displaystyle {\begin{aligned}H_{0}(x)&=1\\H_{1}(x)&=2x\\H_{2}(x)&=4x^{2}-2\end{aligned}}} (2.3)

Verify that the equations in (2.3) are homogeneous solutions of the Hermite differential equation (Eq. 2.2).

## Solution

- Solved on our own


### Part 1

Legendre equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.4).

 {\displaystyle \displaystyle {\begin{aligned}G(x,y,y^{\prime },y^{''})&=g(x,y,p)+f(x,y,p)y^{''}\\&=\underbrace {(1-x^{2})} _{f(x,y,p)}y^{''}\underbrace {-2xy^{\prime }+n(n+1)y} _{g(x,y,p)}\end{aligned}}} (2.4)

Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

 {\displaystyle \displaystyle {\begin{aligned}&f_{xx}=-2\\&f_{xy}=0\\&f_{yy}=0\\&g_{xp}=-2\\&g_{yp}=0\\&g_{y}=n(n+1)\\&f_{xp}=0\\&f_{y}=0\\&f_{yp}=0\\&g_{pp}=0\end{aligned}}} (2.5)
 {\displaystyle \displaystyle {\begin{aligned}f_{xx}+2pf_{xy}+p^{2}f_{yy}&=g_{xp}+pg_{yp}-g_{y}\\-2+2p\cdot 0+p^{2}\cdot 0&=-2+p\cdot 0-n(n+1)\\\end{aligned}}} (2.6)
 {\displaystyle \displaystyle {\begin{aligned}f_{xp}+pf_{yp}+2f_{y}&=g_{pp}\\0+p\cdot 0+2\cdot 0&=0\end{aligned}}} (2.7)
- The second exactness condition is satisfied when ${\displaystyle \displaystyle n=0}$ or ${\displaystyle \displaystyle n=-1}$.


Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

 ${\displaystyle \displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0}$ (2.8)

Using the definition of ${\displaystyle \displaystyle g_{i}:={\frac {\partial G}{\partial y^{(i)}}}}$, the followings are computed.

 {\displaystyle \displaystyle {\begin{aligned}g_{0}&={\frac {\partial G}{\partial y^{(0)}}}=n(n+1)\\g_{1}&={\frac {\partial G}{\partial y^{(1)}}}=-2x\\g_{2}&={\frac {\partial G}{\partial y^{(2)}}}=1-x^{2}\\&{\frac {dg_{1}}{dx}}=-2\\&{\frac {d^{2}g_{2}}{dx^{2}}}=-2\end{aligned}}} (2.9)

After substituting the terms in Eq. (2.9), Eq. (2.8) becomes as following.

 {\displaystyle \displaystyle {\begin{aligned}n(n+1)-(-2)-2&=0\\n(n+1)&=\end{aligned}}} (2.10)
- The second exactness condition is satisfied when ${\displaystyle \displaystyle n=0}$ or ${\displaystyle \displaystyle n=-1}$.


Hermite equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.8).

 {\displaystyle \displaystyle {\begin{aligned}G(x,y,y^{\prime },y^{''})&=g(x,y,p)+f(x,y,p)y^{''}\\&=\underbrace {1} _{f(x,y,p)}\cdot y^{''}\underbrace {-2xy^{\prime }+2ny} _{g(x,y,p)}\end{aligned}}} (2.11)

Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

 {\displaystyle \displaystyle {\begin{aligned}&f_{xx}=0\\&f_{xy}=0\\&f_{yy}=0\\&g_{xp}=-2\\&g_{yp}=0\\&g_{y}=2n\\&f_{xp}=0\\&f_{y}=0\\&f_{yp}=0\\&g_{pp}=0\end{aligned}}} (2.12)
 {\displaystyle \displaystyle {\begin{aligned}f_{xx}+2pf_{xy}+p^{2}f_{yy}&=g_{xp}+pg_{yp}-g_{y}\\0+2p\cdot 0+p^{2}\cdot 0&=-2+p\cdot 0-2n\\\end{aligned}}} (2.13)
 {\displaystyle \displaystyle {\begin{aligned}f_{xp}+pf_{yp}+2f_{y}&=g_{pp}\\0+p\cdot 0+2\cdot 0&=0\end{aligned}}} (2.14)
- The second exactness condition is satisfied only when ${\displaystyle \displaystyle n=-1}$, but the condition is not satisfied if ${\displaystyle \displaystyle n\neq -1}$.


Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

 ${\displaystyle \displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0}$ (2.15)

Using the definition of ${\displaystyle \displaystyle g_{i}:={\frac {\partial G}{\partial y^{(i)}}}}$, the followings are computed.

 {\displaystyle \displaystyle {\begin{aligned}g_{0}&={\frac {\partial G}{\partial y^{(0)}}}=2n\\g_{1}&={\frac {\partial G}{\partial y^{(1)}}}=-2x\\g_{2}&={\frac {\partial G}{\partial y^{(2)}}}=1\\&{\frac {dg_{1}}{dx}}=-2\\&{\frac {d^{2}g_{2}}{dx^{2}}}=0\end{aligned}}} (2.16)

After substituting the terms in Eq. (2.16), Eq. (2.15) becomes as following.

 {\displaystyle \displaystyle {\begin{aligned}2n-(-2)+0&=0\\2(n+1)&=\end{aligned}}} (2.17)
- The second exactness condition is satisfied only when ${\displaystyle \displaystyle n=-1}$.


### Part 2

As shown in the Eq. (2.10), the Hermite equation does not satisfy the second exactness condition if n is a positive value. To see if the equation can be made exact using IFM with ${\displaystyle \displaystyle h(x,y)=x^{m}y^{n}}$, multiply the Eq. (2.2) by ${\displaystyle \displaystyle h(x,y)}$. Also, the term n in the Eq. (2.2) is replaced with ${\displaystyle \displaystyle \lambda }$ to avoid confusion.

 ${\displaystyle \displaystyle x^{m}y^{n}(y^{''}-2xy^{\prime }+2\lambda y)=0}$ (2.18)
 ${\displaystyle \displaystyle \underbrace {x^{m}y^{n}} _{f(x,y,p)}y^{''}\underbrace {-2x^{m+1}y^{n}y^{\prime }+2\lambda x^{m}y^{n+1}} _{g(x,y,p)}=0}$ (2.19)

It is assumed that the Eq. (2.13) is exact. Then, the integrating factor ${\displaystyle \displaystyle h(x,y)}$ can be obtained by utilizing the second exactness condition as following.

 {\displaystyle \displaystyle {\begin{aligned}&f_{xx}=m(m-1)x^{m-2}y^{n}\\&f_{xy}=mnx^{m-1}y^{n-1}\\&f_{yy}=n(n-1)x^{m}y^{n-2}\\&g_{xp}=-2(m+1)x^{m}y^{n}\\&g_{yp}=-2nx^{m+1}y^{n-1}\\&g_{y}=-2nx^{m+1}y^{n-1}p+2(n+1)\lambda x^{m}y^{n}\\&f_{xp}=0\\&f_{y}=nx^{m}y^{n-1}\\&f_{yp}=0\\&g_{pp}=0\end{aligned}}} (2.20)

With the terms computed above, the two equations below which represent the second exactness condition are exploited.

 ${\displaystyle \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}}$ (2.21)
 ${\displaystyle \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}}$ (2.22)

First, substitute the terms in the Eq. (2.14) into the Eq. (2.16).

 {\displaystyle \displaystyle {\begin{aligned}f_{xp}+pf_{yp}+2f_{y}&=g_{pp}\\0+p\cdot 0+2nx^{m}y^{n-1}&=0\end{aligned}}} (2.23)

The only way the Eq. (2.17) is satisfied is to set n equal to zero.

 ${\displaystyle \displaystyle \therefore n=0}$ (2.24)

With knowing n=0, substitute the terms in the Eq. (2.14) into the Eq. (2.15).

 {\displaystyle \displaystyle {\begin{aligned}f_{xx}+2pf_{xy}+p^{2}f_{yy}&=g_{xp}+pg_{yp}-g_{y}\\m(m-1)x^{m-2}&=-2(m+1)x^{m}-2\lambda x^{m}\\&=-2\left(m+1+\lambda \right)x^{m}\end{aligned}}} (2.25)

The both sides of the Eq. (2.19) are equal only when the coefficients of ${\displaystyle \displaystyle x^{m-1},x^{m}}$ are zero. Then the following is concluded.

 {\displaystyle \displaystyle {\begin{aligned}m(m-1)&=0\\m+1+\lambda &=0\end{aligned}}} (2.26)

The first equation in the Eq. (2.20) implies that ${\displaystyle \displaystyle m=0}$ or ${\displaystyle \displaystyle m=1}$. Then ${\displaystyle \displaystyle \lambda }$ is determined using the second one in the Eq. (2.20).

 {\displaystyle \displaystyle {\begin{aligned}&\lambda =-1{\mbox{ when }}m=0\\&\lambda =-2{\mbox{ when }}m=1\end{aligned}}} (2.27)

Since n is determined to be zero already, m can not be zero. Hence, m is chosen to be one and ${\displaystyle \displaystyle \lambda }$ is 2.

 {\displaystyle \displaystyle {\begin{aligned}n&=0\\m&=1\\\lambda &=-2\end{aligned}}} (2.28)
 {\displaystyle \displaystyle {\begin{aligned}\therefore h(x,y)&=x^{1}y^{0}\\&=x\end{aligned}}} (2.29)

Here, the exactness of the Hermite equation with ${\displaystyle \displaystyle h(x,y)}$ is justified.

 {\displaystyle \displaystyle {\begin{aligned}G(x,y,y^{\prime },y^{''})&=g(x,y,p)+f(x,y,p)y^{''}\\&=xy^{''}-2x^{2}y^{\prime }-4xy\end{aligned}}} (2.30)

In order to prove the second exactness of the Hermite equation, the Method 2 is used.

 ${\displaystyle \displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0}$ (2.31)

Using the definition of ${\displaystyle \displaystyle g_{i}:={\frac {\partial G}{\partial y^{(i)}}}}$, the followings are computed.

 {\displaystyle \displaystyle {\begin{aligned}g_{0}&={\frac {\partial G}{\partial y^{(0)}}}=-4x\\g_{1}&={\frac {\partial G}{\partial y^{(1)}}}=-2x^{2}\\g_{2}&={\frac {\partial G}{\partial y^{(2)}}}=x\\&{\frac {dg_{1}}{dx}}=-4x\\&{\frac {d^{2}g_{2}}{dx^{2}}}=0\end{aligned}}} (2.32)

After substituting the terms in Eq. (2.32), Eq. (2.31) becomes as following.

 ${\displaystyle \displaystyle -4x-(-4x)+0=0}$ (2.33)
- Hermite equation with IFM ${\displaystyle \displaystyle h(x,y)=x}$ is exact.


### Part 3

Hermite differential equation is known to have a polynomial solution, which has following form.

 ${\displaystyle \displaystyle y(x)=\sum _{n=0}^{\infty }a_{n}x^{n}}$ (2.34)

Compute the first and the second derivative of Eq. (2.34) with respect to x,

 {\displaystyle \displaystyle {\begin{aligned}y^{\prime }(x)&=\sum _{n=1}^{\infty }na_{n}x^{n-1}\\y^{''}(x)&=\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-2}\\\end{aligned}}} (2.35)

Substitute Eq. (2.34), (2.35) into Eq. (2.2),

 {\displaystyle \displaystyle {\begin{aligned}&\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-1}-2x\sum _{n=1}^{\infty }na_{n}x^{n-1}+2\lambda \sum _{n=1}^{\infty }a_{n}x^{n}=0\\&\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-1}-\sum _{n=1}^{\infty }2na_{n}x^{n}+\sum _{n=1}^{\infty }2\lambda a_{n}x^{n}=0\\\end{aligned}}} (2.36)

Shift the first summation up by two units,

 ${\displaystyle \displaystyle \sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}x^{n}-\sum _{n=1}^{\infty }2na_{n}x^{n}+\sum _{n=0}^{\infty }2\lambda a_{n}x^{n}=0}$ (2.37)

The second summation starts at n=1, while the other summations start at n=0. Following justifies the second summation starting at n=0.

 {\displaystyle \displaystyle {\begin{aligned}\sum _{n=1}^{\infty }2na_{n}x^{n}&=2\cdot 0\cdot a_{0}x^{0}+\sum _{n=1}^{\infty }2na_{n}x^{n}\\&=\sum _{n=0}^{\infty }2na_{n}x^{n}\end{aligned}}} (2.38)

Combine all summations in Eq. (2.37),

 ${\displaystyle \displaystyle \sum _{n=0}^{\infty }\left[(n+2)(n+1)a_{n+2}-2na_{n}+2\lambda a_{n}\right]x^{n}=0}$ (2.39)

Hence, the following recurrence relation is obtained from Eq. (2. 39).

 ${\displaystyle \displaystyle (n+2)(n+1)a_{n+2}-2na_{n}+2\lambda a_{n}=0{\mbox{ for all }}n=0,1,2,3,\cdots }$ (2.40)

Simplify Eq. (2.40),

 ${\displaystyle \displaystyle a_{n+2}={\frac {2(n-\lambda )}{(n+2)(n+1)}}a_{n}{\mbox{ for all }}n=0,1,2,3,\cdots }$ (2.41)

Using the recurrence relation in Eq. (2.41), the Hermite polynomials given in the problem are obtained.
Let the initial conditions and ${\displaystyle \displaystyle \lambda }$ be as follows.

 ${\displaystyle \displaystyle y(0)=a_{0}=1,{\mbox{ }}y^{\prime }(0)=a_{1}=0,{\mbox{ }}\lambda =0}$ (2.42)

Then,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}&={\frac {2(0-0)}{2\cdot 1}}a_{0}=0\\a_{3}&={\frac {2(1-0)}{3\cdot 2}}a_{1}=0\\a_{4}&=a_{5}=a_{6}=a_{7}=\ldots =0\end{aligned}}} (2.43)
- Hence, Hermite polynomial is ${\displaystyle \displaystyle H_{0}(x)=1}$ in this case.


Let the initial conditions and ${\displaystyle \displaystyle \lambda }$ be as follows.

 ${\displaystyle \displaystyle y(0)=a_{0}=0,{\mbox{ }}y^{\prime }(0)=a_{1}=2,{\mbox{ }}\lambda =1}$ (2.44)

Then,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}&={\frac {2(0-1)}{2\cdot 1}}a_{0}=0\\a_{3}&={\frac {2(1-1)}{3\cdot 2}}a_{1}=0\\a_{4}&=a_{5}=a_{6}=a_{7}=\ldots =0\end{aligned}}} (2.45)
- Hence, Hermite polynomial is ${\displaystyle \displaystyle H_{1}(x)=2x}$ in this case.


Let the initial conditions and ${\displaystyle \displaystyle \lambda }$ be as follows.

 ${\displaystyle \displaystyle y(0)=a_{0}=-2,{\mbox{ }}y^{\prime }(0)=a_{1}=0,{\mbox{ }}\lambda =2}$ (2.46)

Then,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}&={\frac {2(0-2)}{2\cdot 1}}a_{0}=4\\a_{3}&={\frac {2(1-2)}{3\cdot 2}}a_{1}=0\\a_{4}&={\frac {2(2-2)}{4\cdot 3}}a_{2}=0\\a_{5}&=a_{6}=a_{7}=a_{8}=\ldots =0\end{aligned}}} (2.47)
- Hence, Hermite polynomial is ${\displaystyle \displaystyle H_{2}(x)=4x^{2}-2}$ in this case.


# References

• Hermite Polynomial, WolframMathworld [1]

# Problem R*5.3 Method of undetermined coefficients

From Mtg 30-3

## Given

Given L4-ODE-CC,

${\displaystyle \displaystyle X^{(4)}-K^{4}X=0}$

(3.1)

Assuming,

${\displaystyle \displaystyle X(x)=e^{(rx)}}$

(3.2)

and substituting this in Eq. (3.1), we get 4 solutions for r,

${\displaystyle \displaystyle r_{1,2}=\pm K}$

${\displaystyle \displaystyle r_{3,4}=\pm i\,K}$

(3.3)

## Find

Expressions for X(x) in terms of ${\displaystyle \displaystyle \cos Kx,\ \sin Kx,\ \cosh Kx,\ \sinh Kx}$

## Solution

- Solved on my own


We can write final solution X(x) as,

${\displaystyle \displaystyle X(x)=\sum _{i=1}^{4}C_{i}e^{(r_{i}x)}}$

(3.4)

Substituting all the ri values from Eq. (3.3),

${\displaystyle \displaystyle X(x)=C_{1}e^{Kx}+C_{2}e^{-Kx}+C_{3}e^{iKx}+C_{4}e^{-iKx}}$

(3.5)

Expanding the exponential terms into sin, cos, sinh and cosh terms, we get,

${\displaystyle \displaystyle X(x)=C_{1}\cosh Kx\ +C_{1}\sinh Kx\ +C_{2}\cosh Kx\ -C_{2}\sinh Kx\ +C_{3}\cos Kx\ +iC_{3}\sin Kx\ +C_{4}\cos kx-\ iC_{4}\sin kx}$

(3.6)

for X(x) to be real, imaginary part in above Eq. (3.6) must be real, i.e. (C3 - C4) must be imaginary

${\displaystyle \displaystyle C_{3}-C_{4}=ib_{4}}$

and, (C3 + C4) must be real,

${\displaystyle \displaystyle C_{3}+C_{4}=b_{3}}$

Therefore, C3 and C4 will be given by,

${\displaystyle \displaystyle C_{3}={\frac {1}{2}}(b_{3}+ib_{4})}$

${\displaystyle \displaystyle C_{4}={\frac {1}{2}}(b_{3}-ib_{4})}$

This means C3 and C4 given above are complex conjugates.

Therefore, we get final solution to be,

${\displaystyle \displaystyle X(x)=(C_{1}+C_{2})\cosh Kx+(C_{1}-C_{2})\sinh Kx+b_{3}\cos Kx-b_{4}\sin Kx}$

(3.7)

# Problem R*5.4 - Find ${\displaystyle \displaystyle y_{xxxxx}}$ in terms of the derivatives of y with respect to t

From Mtg 31-1

## Given

In Euler L2-ODE-VC, suppose: ${\displaystyle \displaystyle x=e^{t}}$.

## Find

${\displaystyle \displaystyle y_{xxxxx}}$

## Solution

- Solved on my own


Since:

 ${\displaystyle \displaystyle {\frac {dt}{dx}}=\left({\frac {dx}{dt}}\right)^{-1}=\left({\frac {d(e^{t})}{dt}}\right)^{-1}=e^{-t}}$ (4.1)

So:

 ${\displaystyle \displaystyle y_{x}={\frac {dy}{dx}}={\frac {dy}{dt}}{\frac {dt}{dx}}=e^{-t}y_{t}}$ (4.2)
 {\displaystyle \displaystyle {\begin{aligned}y_{xx}&={\frac {dy_{x}}{dx}}\\&={\frac {dy_{x}}{dt}}{\frac {dt}{dx}}\\&=e^{-2t}(y_{tt}-y_{t})\\\end{aligned}}} (4.3)
 {\displaystyle \displaystyle {\begin{aligned}y_{xxx}&={\frac {dy_{xx}}{dx}}\\&={\frac {dy_{xx}}{dt}}{\frac {dt}{dx}}\\&=e^{-3t}(y_{ttt}-3y_{tt}+2y_{t})\\\end{aligned}}} (4.4)
 {\displaystyle \displaystyle {\begin{aligned}y_{xxxx}&={\frac {dy_{xxx}}{dx}}\\&={\frac {dy_{xxx}}{dt}}{\frac {dt}{dx}}\\&=e^{-4t}(y_{tttt}-6y{ttt}+11y_{tt}-6y_{t})\\\end{aligned}}} (4.5)
 {\displaystyle \displaystyle {\begin{aligned}y_{xxxxx}&={\frac {dy_{xxxx}}{dx}}\\&={\frac {dy_{xxxx}}{dt}}{\frac {dt}{dx}}\\&=e^{-t}{\frac {d\left(e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t})\right)}{dt}}\\&=e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_{t})\end{aligned}}} (4.6)

# Problem R*5.5 - Solve y and plot the solution

From Mtg 31-3

## Given

Euler L2-ODE-VC:

 ${\displaystyle \displaystyle x^{2}y''-2xy'+2y=0}$ (5.1)

and trial solution:

 ${\displaystyle \displaystyle y=x^{r}}$ (5.2)

with boundary conditions:

 ${\displaystyle \displaystyle {\begin{cases}y(1)=-4\\y(2)=7\\\end{cases}}}$ (5.3)

## Find

The solution ${\displaystyle \displaystyle y(x)}$ and plot it.

## Solution

- Solved on my own


Substitute ${\displaystyle \displaystyle x^{r}}$ for ${\displaystyle \displaystyle y}$ in equation (5.1):

 ${\displaystyle \displaystyle \underbrace {x^{r}} _{\displaystyle \neq 0}(r^{2}-3r+2)=0}$ (5.4)

So the characteristic equation:

 ${\displaystyle \displaystyle r^{2}-3r+2=0\Rightarrow {\begin{cases}r_{1}=1\\r_{2}=2\\\end{cases}}}$ (5.5)

So the solution is:

 ${\displaystyle \displaystyle y(x)=c_{1}x+c_{2}x^{2}}$ (5.6)

Use the boundary conditions:

 ${\displaystyle \displaystyle {\begin{cases}y(1)=c_{1}+c_{2}=-4\\y(2)=2c_{1}+4c_{2}=7\\\end{cases}}\Rightarrow {\begin{cases}c_{1}=-{\frac {23}{2}}\\c_{2}={\frac {15}{2}}\\\end{cases}}}$ (5.7)

So the solution is:

 ${\displaystyle \displaystyle y(x)={\frac {15}{2}}x^{2}-{\frac {23}{2}}x}$ (5.8)

Plot ${\displaystyle \displaystyle y(x)}$.

Matlab Code:

 clear all close all clc %homogeneous solutions y=@(x) (15/2)*(x^2)-(23/2)*x; %plotting hold on fplot(y,[-7 7],'r'); title('Plotting'); xlabel('x'); ylabel('y(x)'); legend('y=(15/2)*(x^2)-(23/2)*x') 

# Problem R5.6: Equivalence of Two methods to solve Euler Ln-ODE-VC

From Mtg 31-5

## Given

Given Euler Ln-ODE-VC,

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}x^{i}y^{(i)}=0}$

(6.1)

where, ${\displaystyle \displaystyle y^{(i)}}$ is ith derivative of y w.r.t. x,

Two methods of solving above equation,

Method 1:

Stage 1: Transformation of variables

${\displaystyle \displaystyle x=e^{t}}$

(6.2)

Stage 2: Trial Solution

${\displaystyle \displaystyle y=e^{rt},\ {\color {red}r={\text{constant}}}}$

(6.3)

Method 2: Trial Solution

${\displaystyle \displaystyle y=x^{r},\ {\color {red}r={\text{constant}}}}$

(6.4)

## Find

Show equivalence of methods 1 and 2 shown above in Eq. (6.2), (6.3) and Eq. (6.4) respectively

## Solution

- Solved on our own


Let's take method - 2 shown in Eq. (6.4), first. Differentiating Eq. (6.4) w.r.t. x, we get

${\displaystyle \displaystyle y_{x}=y'=rx^{r-1}}$

(6.5)

For, ith derivative of y w.r.t. x, i.e. y(i), we have,

${\displaystyle \displaystyle y^{(i)}=r(r-1)...(r-i+1)x^{(r-1)}}$

(6.6)

Substituting Eq. (6.6), i.e. definition of y(i) in Eq. (6.1), we get,

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}x^{i}r(r-1)...(r-i+1)x^{(r-i)}=0}$

(6.7a)

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}x^{r}r(r-1)...(r-i+1)=0}$

(6.7b)

Eliminating xr, which is a common factor in the all the terms of above summation, we get,

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}r(r-1)...(r-i+1)=0}$

(6.7c)

So, we solve the above equation for r, and that provides us solution for Eq. (6.1).

Now, for method - 1, we will try to get the same form as in Eq. (6.7c)

Stage 1: Substitute x = et in Eq. (6.1), we write yx as,

${\displaystyle \displaystyle y_{x}=y_{t}{\frac {dt}{dx}}=e^{-t}y_{t}}$

(6.8)

Stage 2: Substitute y = ert,

Using above we can write yt as, r*ert. Substitute this into Eq. (6.8), and we get,

${\displaystyle \displaystyle y^{(1)}=y_{x}=re^{rt}e^{-}t=re^{(r-1)t}}$

(6.9)

Performing another differentiation of above equation to get yxx, we get,

${\displaystyle \displaystyle y^{(2)}=y_{xx}={\frac {d}{dx}}(y_{x})={\frac {d}{dx}}(re^{(r-1)t})={\frac {d}{dt}}(re^{(r-1)t}){\frac {dt}{dx}}=r(r-1)e^{(r-1)t}e^{-t}=r(r-1)e^{(r-2)t}}$

(6.10)

We can see that as done above for y(2), we can write in general for y(i), we can write,

${\displaystyle \displaystyle y^{(i)}=r(r-1)...(r-i+1)e^{(r-i)t}}$

(6.10)

Substituting definitions of both x and y as seen in Eq. (6.2) and Eq. (6.10) respectively, into Eq. (6.1) we get,

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}e^{it}r(r-1)...(r-i+1)e^{(r-i)t}=0}$

(6.11a)

Above can be simplified to following form very easily (combining both the exponential terms in above expression),

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}e^{rt}r(r-1)...(r-i+1)=0}$

(6.11b)

Since, e(rt) is not dependent on i, we can take this term out of the summation as a common factor, and write rest of the terms as,

${\displaystyle \displaystyle \sum _{i=0}^{n}a_{i}r(r-1)...(r-i+1)=0}$

(6.11c)

Above Eq. (6.11c) is exactly same as Eq. (6.7c), and therefore it proves that by both methods 1 & 2 we will get same values for r,
and hence both methods are equivalent.


# Problem R*5.7-Euler L2-ODEs

From Mtg 32-1

## Given

Characteristic equation

 {\displaystyle \displaystyle {\begin{aligned}(r-\lambda )^{2}=0\end{aligned}}} (7.1)
 {\displaystyle \displaystyle {\begin{aligned}a_{2}x^{2}y''+a_{1}xy'+a_{0}y=0\end{aligned}}} (7.2)
 {\displaystyle \displaystyle {\begin{aligned}b_{2}y''+b_{1}y'+b_{0}y=0\end{aligned}}} (7.3)

## Find

1.1) Find ${\displaystyle a_{2},a_{1},a_{0}}$ such that (7.1) is characteristic equation of (7.2)

1.2) 1st homogeneous solution : ${\displaystyle y_{1}(x)=x^{\lambda }}$

1.3) Complete solution : Find c(x) such that ${\displaystyle y(x)=c(x)y_{1}(x)}$

1.4) Find the 2nd homogeneous solution ${\displaystyle y_{2}(x)}$

2.1) Find ${\displaystyle b_{2},b_{1},b_{0}}$ such that (7.1) is characteristic equation of (7.3)

2.2) 1st homogeneous solution

2.3) Complete solution : Find c(x) such that ${\displaystyle y(x)=c(x)y_{1}(x)}$

2.4) Find the 2nd homogeneous solution ${\displaystyle y_{2}(x)}$

## Solution

- Solved on our own


### Part 1.1

Assume that the trial solution is following.

 {\displaystyle \displaystyle {\begin{aligned}y=x^{r}\end{aligned}}} (7.4)

Then, compute the first and the second derivatives of the trial solution,

 {\displaystyle \displaystyle {\begin{aligned}y'=rx^{r-1}\end{aligned}}} (7.5)
 {\displaystyle \displaystyle {\begin{aligned}y''=r(r-1)x^{r-2}\end{aligned}}} (7.6)

Substitute the equation (7.4), (7.5), (7.6) into (7.2) and simplify the equation,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}x^{2}r(r-1)x^{r-2}+a_{1}xrx^{r-1}+a_{0}x^{r}=0\\a_{2}r(r-1)x^{r}+a_{1}rx^{r}+a_{0}x^{r}=0\\\left[a_{2}r(r-1)+a_{1}r+a_{0}\right]x^{r}=0\end{aligned}}} (7.7)

Then, we get

 {\displaystyle \displaystyle {\begin{aligned}a_{2}r^{2}+(a_{1}-a_{2})r+a_{0}=0\end{aligned}}} (7.8)

Compare (7.8) to (7.1),

 {\displaystyle \displaystyle {\begin{aligned}a_{2}r^{2}+(a_{1}-a_{2})r+a_{0}&=(r-\lambda )^{2}\\&=r^{2}-2r\lambda +\lambda ^{2}\\\end{aligned}}} (7.9)

Comparing the coefficients of LHS and RHS,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}&=1\\a_{1}&=1-2\lambda \\a_{0}&=\lambda ^{2}\end{aligned}}} (7.10)

### Part 1.2

From the characteristic equation in Eq. (7.1), the solution of the characteristic equation is determined as follows.

 ${\displaystyle \displaystyle r=\lambda }$ (7.11)

Hence, the 1st homogeneous solution can be

 {\displaystyle \displaystyle {\begin{aligned}y_{1}=x^{\lambda }\end{aligned}}} (7.12)

### Part 1.3

 {\displaystyle \displaystyle {\begin{aligned}y=c(x)y_{1}(x)\end{aligned}}} (7.13)

Compute the first and the second derivatives of Eq. (7.13),

 {\displaystyle \displaystyle {\begin{aligned}y^{\prime }(x)&=c^{\prime }(x)y_{1}(x)+c(x)y_{1}^{\prime }(x)\\y^{''}(x)&=c^{''}(x)y_{1}(x)+c^{\prime }(x)y_{1}^{\prime }(x)+c^{\prime }(x)y_{1}^{\prime }(x)+C(x)y_{1}^{''}(x)\end{aligned}}} (7.14)

Substitute (7.13) and (7.14) into (7.2) and simplify the equation,

 {\displaystyle \displaystyle {\begin{aligned}a_{2}x^{2}\left[c^{''}(x)y_{1}(x)+c^{\prime }(x)y_{1}^{\prime }(x)+c^{\prime }(x)y_{1}^{\prime }(x)+c(x)y_{1}^{''}(x)\right]+a_{1}x\left[c^{\prime }(x)y_{1}(x)+c(x)y_{1}^{\prime }(x)\right]+a_{0}c(x)y_{1}(x)&=0\\c(x){\cancelto {0}{\left[a_{2}x^{2}y_{1}^{''}(x)+a_{1}xy_{1}^{\prime }(x)+a_{0}y_{1}(x)\right]}}+c^{\prime }(x)\left[2a_{2}x^{2}y_{1}^{\prime }(x)+a_{1}xy_{1}(x)\right]+c^{''}(x)\left[a_{2}x^{2}y_{1}(x)\right]&=0\\\end{aligned}}} (7.15)

The first term in Eq. (7.15) cancels out using Eq. (7.2). Then,

 {\displaystyle \displaystyle {\begin{aligned}c^{\prime }(x)\left[2a_{2}x^{2}y_{1}^{\prime }(x)+a_{1}xy_{1}(x)\right]+c^{''}(x)\left[a_{2}x^{2}y_{1}(x)\right]=0\\\end{aligned}}} (7.16)

Substitute coefficients and homogeneous solution,

 {\displaystyle \displaystyle {\begin{aligned}c^{\prime }(x)\left[2a_{2}x^{2}y_{1}^{\prime }(x)+a_{1}xy_{1}(x)\right]+c^{''}(x)\left[a_{2}x^{2}y_{1}(x)\right]\\\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}&=c^{\prime }(x)\left[2x^{2}\lambda x^{\lambda -1}+1-2\lambda xx^{\lambda }\right]+c^{''}(x)\left[x^{2}x^{\lambda }\right]\\&=c^{\prime }(x)\left[2\lambda x^{\lambda +1}+1-2\lambda x^{\lambda +1}\right]+c^{''}(x)\left[x^{\lambda +2}\right]\\&=c^{\prime }(x)x^{\lambda +1}+c^{''}(x)x^{\lambda +2}\\&=\left[c^{\prime }(x)+c^{''}(x)x\right]x^{\lambda +1}=0\\\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}\therefore \left[c^{\prime }(x)+c^{''}(x)x\right]=0\end{aligned}}} (7.17)

Let ${\displaystyle \displaystyle c^{\prime }(x)=z(x)}$,

 {\displaystyle \displaystyle {\begin{aligned}z(x)+z'(x)x&=0\\z+x{\frac {dz}{dx}}&=0\\z&=-x{\frac {dz}{dx}}\\{\frac {1}{z}}dz&=-{\frac {1}{x}}dx\\\int {\frac {1}{z}}dz&=\int {\frac {1}{x}}dx\\\end{aligned}}} (7.18)
 ${\displaystyle \displaystyle \log(z)=\log(x)+C_{1}}$ (7.19)
 {\displaystyle \displaystyle {\begin{aligned}z&=e^{-(\log(x)+C_{1})}\\&=e^{\log(x{e^{C_{1}}})^{-1}}\\&={\frac {1}{e^{C_{1}}x}}\\&=C_{2}{\frac {1}{x}}{\mbox{ }}(C_{2}=e^{-C_{1}})\end{aligned}}} (7.20)

Then, compute the function ${\displaystyle \displaystyle c(x)}$,

 {\displaystyle \displaystyle {\begin{aligned}c(x)&=\int zdx\\&=\int C_{2}{\frac {1}{x}}dx\\&=C_{2}(\log(x)+C_{3})\end{aligned}}} (7.21)

As a result,

 {\displaystyle \displaystyle {\begin{aligned}y(x)&=C_{2}(\log(x)+C_{3})y_{1}(x)\\&=C_{2}x^{\lambda }\log(x)+C_{4}x^{\lambda }\\&=C_{1}^{*}x^{\lambda }\log(x)+C_{2}^{*}x^{\lambda }\end{aligned}}} (7.22)

### Part 1.4

From 1.3, the general homogeneous solution is

 {\displaystyle \displaystyle {\begin{aligned}y(x)&=C_{1}^{*}x^{\lambda }\log(x)+C_{2}^{*}x^{\lambda }\end{aligned}}}

From (7.12), ${\displaystyle \displaystyle y_{1}(x)}$ is ${\displaystyle \displaystyle x^{\lambda }}$.

So, ${\displaystyle \displaystyle y_{2}(x)}$ is

 {\displaystyle \displaystyle {\begin{aligned}y_{2}(x)&=x^{\lambda }\log(x)\end{aligned}}}

### Part 2.1

The trial solution for the Euler equation with constant coefficient:

 {\displaystyle \displaystyle {\begin{aligned}y=e^{xr}\end{aligned}}} (7.23)

Compute the first and the second derivatives

 {\displaystyle \displaystyle {\begin{aligned}y^{\prime }(x)=re^{xr}\end{aligned}}} (7.24)
 {\displaystyle \displaystyle {\begin{aligned}y^{''}(x)=r^{2}e^{rx}\end{aligned}}} (7.25)

Substitute (7.23),(7.24),(7.25) into (7.3) and simplify it,

 {\displaystyle \displaystyle {\begin{aligned}b_{2}r^{2}e^{rx}+b_{1}re^{rx}+b_{0}e^{rx}&=0\\\left[b_{2}r^{2}+b_{1}r+b_{0}\right]e^{rx}&=0\\\end{aligned}}}

Then,

 {\displaystyle \displaystyle {\begin{aligned}b_{2}r^{2}+b_{1}r+b_{0}=0\\\end{aligned}}}
 {\displaystyle \displaystyle {\begin{aligned}b_{2}r^{2}+b_{1}r+b_{0}&=(r-\lambda )^{2}\\&=r^{2}-2r\lambda +\lambda ^{2}\end{aligned}}}

Compare LHS and RHS,

 {\displaystyle \displaystyle {\begin{aligned}b_{2}=1\end{aligned}}} (7.26)
 {\displaystyle \displaystyle {\begin{aligned}b_{1}=-2\lambda \end{aligned}}} (7.27)
 {\displaystyle \displaystyle {\begin{aligned}b_{0}=\lambda ^{2}\end{aligned}}} (7.28)

### Part 2.2

From (7.1), ${\displaystyle \displaystyle r=\lambda }$, As a result,

 {\displaystyle \displaystyle {\begin{aligned}y_{1}=e^{x\lambda }\end{aligned}}} (7.29)

### Part 2.3

 {\displaystyle \displaystyle {\begin{aligned}y=c(x)y_{1}(x)\end{aligned}}} (7.30)

Compute the first and the second derivatives

 {\displaystyle \displaystyle {\begin{aligned}y^{'}(x)=c^{'}(x)y_{1}(x)+c(x)y_{1}^{'}(x)\end{aligned}}} (7.31)
 {\displaystyle \displaystyle {\begin{aligned}y^{''}(x)=c^{''}(x)y_{1}(x)+c^{'}(x)y_{1}^{'}(x)+c^{'}(x)y_{1}^{'}(x)+c(x)y_{1}^{''}(x)\end{aligned}}} (7.32)

Substituting (7.25),(7.26) and (7.27) into (7.3)

 {\displaystyle \displaystyle {\begin{aligned}b_{2}\left[c^{''}(x)y_{1}(x)+c^{'}(x)y_{1}^{'}(x)+c^{'}(x)y_{1}^{'}(x)+C(x)y_{1}^{''}(x)\right]+b_{1}\left[c^{'}(x)y_{1}(x)+C(x)y_{1}^{'}(x)\right]+b_{0}c(x)y_{1}(x)\\c(x){\cancelto {0}{\left[b_{2}y_{1}^{''}(x)+b_{1}y_{1}^{'}(x)+b_{0}y_{1}(x)\right]}}+c^{'}(x)\left[2b_{2}y_{1}^{'}(x)+b_{1}y_{1}(x)\right]+c^{''}(x)\left[b_{2}y_{1}(x)\right]=0\\\end{aligned}}}

From (7.3),

 {\displaystyle \displaystyle {\begin{aligned}b_{2}y_{1}^{''}(x)+b_{1}y_{1}^{'}(x)+b_{0}y_{1}(x)=0\end{aligned}}}

Then,

 {\displaystyle \displaystyle {\begin{aligned}c^{'}(x)\left[2b_{2}y_{1}^{'}(x)+b_{1}y_{1}(x)\right]+c^{''}(x)\left[b_{2}y_{1}(x)\right]=0\\\end{aligned}}}

Substituting coefficients and homogeneous solution,

 {\displaystyle \displaystyle {\begin{aligned}&c^{'}(x)\left[2b_{2}y_{1}^{'}(x)+b_{1}y_{1}(x)\right]+c^{''}(x)\left[b_{2}y_{1}(x)\right]\\&=c^{'}(x){\cancelto {0}{\left[2\lambda e^{\lambda x}-2\lambda e^{\lambda x}\right]}}+c^{''}(x)\left[e^{\lambda x}\right]\\&=c^{''}(x)e^{\lambda x}=0\\\end{aligned}}}

Then,

 {\displaystyle \displaystyle {\begin{aligned}\therefore c^{''}(x)=0\end{aligned}}} (7.33)

Solve (7.33),

 {\displaystyle \displaystyle {\begin{aligned}c^{'}(x)&=\int c^{''}(x)dx\\&=\int 0dx\\&=C_{1}\\c^{'}(x)&=C_{1}\\c(x)&=\int c^{'}(x)dx\\&=\int C_{1}dx\\&=C_{1}x+C_{2}\end{aligned}}} (7.34)

As a result,

 {\displaystyle \displaystyle {\begin{aligned}y(x)&=c(x)y_{1}(x)\\&=(C_{1}x+C_{2})e^{\lambda x}\\&=C_{1}xe^{\lambda x}+C_{2}e^{\lambda x}\end{aligned}}} (7.35)

### Part 2.4

From 2.3, the general homogeneous solution is

 {\displaystyle \displaystyle {\begin{aligned}y(x)&=(C_{1}x+C_{2})e^{\lambda x}\\&=C_{1}xe^{\lambda x}+C_{2}e^{\lambda x}\\\end{aligned}}}

From (7.29), ${\displaystyle \displaystyle y_{1}(x)}$ is ${\displaystyle \displaystyle e^{\lambda x}}$.

So, ${\displaystyle \displaystyle y_{2}(x)}$ is

 {\displaystyle \displaystyle {\begin{aligned}y_{2}(x)&=xe^{\lambda x}\end{aligned}}}

# Problem R*5.8 - Variation of parameters

From Mtg 32-2

## Given

A General non-homogenous L1-ODE-VC (Linear, 1st order, Ordinary Differential Equation with Variable Coefficient),

 ${\displaystyle \displaystyle P(x)\,y'+Q(x)\,y=\underbrace {R(x)} _{\color {blue}{\neq 0}}}$ (8.1)

which can be written in the form given below,

 ${\displaystyle \displaystyle \underbrace {\color {blue}{1}} _{\color {blue}{a_{1}(x)}}\cdot \underbrace {y'} _{\color {blue}{y^{(1)}}}+\underbrace {\frac {Q(x)}{P(x)}} _{\color {blue}{a_{0}(x)}}y^{\color {blue}{(0)}}=\underbrace {\frac {R(x)}{P(x)}} _{\color {blue}{b(x)}}}$ (8.2)

The form can be written as,

 ${\displaystyle \displaystyle y'+a_{0}(x)y^{(0)}=b(x)}$ (8.3)

## Find

Find the particular solution, ${\displaystyle \displaystyle y_{P}(x)}$ by the use of the method of variation of parameters after knowing the homogeneous solution ${\displaystyle \displaystyle y_{H}(x)}$, i.e., let ${\displaystyle \displaystyle y(x)=A(x)y_{H}(x)}$, with ${\displaystyle \displaystyle A(x)}$ being the unknown to be found.

## Solution

- Solved on our own


For particular solution, ${\displaystyle \displaystyle y_{P}(x)}$, the use variation of parameters requires ${\displaystyle \displaystyle y_{P}(x)}$ to be following,

 ${\displaystyle \displaystyle y_{p}(x)=\sum _{i=1}^{n}c_{i}(x)y_{i}(x)}$ (8.4)

for ${\displaystyle n^{th}}$ order non-linear homogenous ODE. For our case, n = 1, therefore we get,

 ${\displaystyle \displaystyle y_{p}(x)=A(x)y_{H}(x)}$ (8.5)

It is important to note for method of variation of parameters that above, ${\displaystyle \displaystyle c_{1}}$ is a function of ${\displaystyle \displaystyle x}$, and is determined by substituting the particular solution. Thus,

 ${\displaystyle \displaystyle y_{H}^{'}(x)+y_{P}^{'}(x)+a_{0}(x)[y_{H}(x)+y_{P}(x)]=b(x)}$ (8.6)

Substitute Eq. 8.5 into Eq. 8.6 to get,

 ${\displaystyle \displaystyle y_{H}^{'}(x)+[A^{'}(x)y_{H}(x)+A(x)y_{H}^{'}(x)]+a_{0}(x)y_{H}(x)+a_{0}(x)[y_{H}(x)+A(x)y_{H}(x)]=b(x)}$ (8.7)

and, rearrange the terms,

 ${\displaystyle \displaystyle \underbrace {[y_{H}^{'}(x)+a_{0}(x)y_{H}(x)]} _{=0}+A^{'}(x)y_{H}(x)+A(x)\underbrace {[y_{H}^{'}(x)+a_{0}(x)y_{H}(x)]} _{=0}=b(x)}$ (8.8)

Therefore ${\displaystyle \displaystyle A(x)}$ is given by,

 ${\displaystyle \displaystyle A^{'}(x)={\frac {b(x)}{y_{H}(x)}}}$ (8.9)

Integrate both sides w.r.t. x,

 ${\displaystyle \displaystyle A(x)=\int ^{x}{\frac {b(s)}{y_{H}(s)}}ds}$ (8.10)

Substituting for the expression of ${\displaystyle \displaystyle y_{H}(x)}$ from the homework problem 2.17 into above, we will get,

 ${\displaystyle \displaystyle A(x)=\int ^{x}\underbrace {\exp[\int ^{s}a_{0}(t)dt]} _{=h(s)}b(s)ds}$ (8.11)

Therefore particular solution can finally be written as,

 ${\displaystyle \displaystyle y_{p}(x)=[\int ^{x}h(s)b(s)ds]{\frac {1}{h(s)}}}$ (8.12)

# Problem R*5.9 Special IFM Solution

From Mtg 32-7

## Given

 {\displaystyle \displaystyle {\begin{aligned}a_{2}x^{2}y''+a_{1}xy'+a_{0}y=0\end{aligned}}} (9.1)

## Find

Special IFM to solve for general ${\displaystyle \displaystyle f(t)}$

## Solution

- Solved on our own


First, find integrating factor ${\displaystyle \displaystyle h(t,y)}$

 ${\displaystyle \displaystyle h(t,y)[{a_{2}}y''+{a_{1}}y'+{a_{0}}y]=h(t,y)f(t)}$ (Eq 9.2)

Suppose ${\displaystyle \displaystyle h(t,y)}$ is a function of ${\displaystyle \displaystyle t}$, thus ${\displaystyle \displaystyle h(t,y)=h(t)}$

Rearrange Eq 9.2, we have

 ${\displaystyle \displaystyle h(t)[{a_{2}}y''+{a_{1}}y'+{a_{0}}y-f(t)]=0}$ (Eq 9.3)
 ${\displaystyle \displaystyle {\frac {d\phi }{dt}}=0}$ (Eq 9.4)

Thus

 ${\displaystyle \displaystyle F(t,y,p,p')=h(t){a_{2}}p'+h(t){a_{1}}p+h(t){a_{0}}y-h(t)f(t)=f(t,y,p)p'+g(t,y,p)}$ (Eq 9.5)

Where ${\displaystyle \displaystyle p:=y'}$

According to the 2nd Exactness Condition

 ${\displaystyle \displaystyle {f}_{tt}+2p{f}_{ty}+{p}^{2}{f}_{yy}={g}_{tp}+p{g}_{yp}-{g}_{y}}$ (Eq 9.6)
 ${\displaystyle \displaystyle {f}_{tp}+p{f}_{yp}+2{f}_{y}={g}_{pp}}$ (Eq 9.7)

We have

 ${\displaystyle \displaystyle f_{tt}={h_{tt}}{a_{2}}}$ (Eq 9.8)
 ${\displaystyle \displaystyle f_{ty}=0}$ (Eq 9.9)
 ${\displaystyle \displaystyle f_{yy}=0}$ (Eq 9.10)
 ${\displaystyle \displaystyle g_{tp}={h_{t}}{a_{1}}}$ (Eq 9.11)
 ${\displaystyle \displaystyle g_{yp}=0}$ (Eq 9.12)
 ${\displaystyle \displaystyle g_{y}={h}{a_{0}}}$ (Eq 9.13)

Substituting Eq 9.8~9.13 in Eq 9.6, we get

 ${\displaystyle \displaystyle h_{tt}a_{2}={h_{t}}{a_{1}}-h{a_{0}}}$ (Eq 9.14)

Now we assume ${\displaystyle \displaystyle h(t)=e^{\alpha t}}$, where ${\displaystyle \displaystyle {\alpha }}$ is a constant

So Eq 9.14 can be expressed as

 ${\displaystyle \displaystyle {\alpha }^{2}{a_{2}}-{\alpha }{a_{1}}+{a_{0}}=0}$ (Eq 9.15)

Then we try to reduce order for Eq 9.2 and express Eq 9.2 as

 ${\displaystyle \displaystyle {{e}^{\alpha t}}\left({{\bar {a}}_{1}}{y}'+{{\bar {a}}_{2}}y\right)=\int {{{e}^{\alpha t}}f(t)dt}}$ (Eq 9.16)

We can get

 ${\displaystyle \displaystyle {{e}^{\alpha t}}\left[{{\bar {a}}_{1}}{y}''+(\alpha {{\bar {a}}_{1}}+{{\bar {a}}_{0}}){y}'+\alpha {{\bar {a}}_{0}}y\right]={{e}^{\alpha t}}f(t)}$ (Eq 9.17)

Therefore, we have

 {\displaystyle \displaystyle {\begin{aligned}&{{\bar {a}}_{1}}={{a}_{2}}\\&{{\bar {a}}_{0}}={{a}_{1}}-\alpha {{a}_{2}}\\&{{\bar {a}}_{0}}={{a}_{0}}/\alpha \\\end{aligned}}} (Eq 9.18)

Rearrange Eq 9.16, we have

 ${\displaystyle \displaystyle \left({y}'+{\frac {{\bar {a}}_{2}}{{\bar {a}}_{1}}}y\right)={\frac {{e}^{-\alpha t}}{{\bar {a}}_{1}}}\int {{{e}^{\alpha t}}f(t)dt}}$ (Eq 9.19)

Using ${\displaystyle \displaystyle h(t)=e^{{\beta }t}}$, where ${\displaystyle \displaystyle {\beta }={\frac {\bar {a_{0}}}{\bar {a_{1}}}}}$

Thus, we can get

 {\displaystyle \displaystyle {\begin{aligned}y(t)&={\frac {{e}^{-\beta t}}{{\bar {a}}_{1}}}\int _{}^{t}{{{e}^{(\beta -\alpha )\tau }}(\int _{}^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\&={\frac {{e}^{-\beta t}}{{\bar {a}}_{1}}}\int _{}^{t}{{{e}^{(\beta -\alpha )\tau }}\left(\int {{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}}\right)d\tau }\\&={\frac {{e}^{-\beta t}}{{\bar {a}}_{1}}}\left[\int _{}^{t}{{{e}^{(\beta -\alpha )\tau }}\left(\int {{{e}^{\alpha \tau }}f(\tau )d\tau }\right)d\tau }+\int _{}^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau }\right]\\&={\frac {{e}^{-\beta t}}{{\bar {a}}_{1}}}\left[\int {{{e}^{(\beta -\alpha )t}}\left(\int {{{e}^{\alpha t}}f(t)dt}\right)dt}+{{k}_{2}}+{\frac {{k}_{1}}{\beta -\alpha }}{{e}^{(\beta -\alpha )t}}\right]\\&={\frac {{k}_{1}}{(\beta -\alpha ){{\bar {a}}_{1}}}}{{e}^{-\alpha t}}+{\frac {{k}_{2}}{{\bar {a}}_{1}}}{{e}^{-\beta t}}+{\frac {{e}^{-\beta t}}{{\bar {a}}_{1}}}\int {{{e}^{(\beta -\alpha )t}}\left(\int {{{e}^{\alpha t}}f(t)dt}\right)dt}\end{aligned}}} (Eq 9.20)

Therefore, we have

 ${\displaystyle \displaystyle y(t)={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+{\frac {{e}^{-\beta t}}{{a}_{2}}}\int {{{e}^{(\beta -\alpha )t}}\left(\int {{{e}^{\alpha t}}f(t)dt}\right)dt}}$ (Eq 9.21)

Where ${\displaystyle \displaystyle {C_{1}}}$ and ${\displaystyle \displaystyle {C_{2}}}$ are constants.

# Contributing Members

 Team Contribution Table Problem Number Lecture(Mtg) Assigned To Solved By Typed By Signature(Author) R*5.1 Mtg 22-6 Chung Chung Chung Chung 17 Oct. 2011 at 1:30 (UTC) R*5.2 Mtg 27-1 Shin Shin Shin Shin 31 Oct. 2011 at 14:38 (UTC) R*5.3 Mtg 30-3 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 18:00 (UTC) R*5.4 Mtg 31-1 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 08:47 (UTC) R*5.5 Mtg 31-3 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 10:01 (UTC) R5.6 Mtg 31-5 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 20:00 (UTC) R*5.7 Mtg 32-1 Chung Chung Chung Chung 29 Oct. 2011 at 0:19 (UTC) R*5.8 Mtg 32-2 Shin Shin Shin Shin 2 Nov. 2011 at 15:43 (UTC) R*5.9 Mtg 32-7 Ren Ren Ren Ren 1 Nov. 2011 at 23:50 (UTC)