# R*2.1 - Homogenous Solutions of the Second-Order Legendre Equation

## Given

The Legendre Differential Equation[1] is a second-order ordinary differential equation given as follows:

 $\displaystyle L_2(y) := (1-x^2)y'' - 2xy' + n(n+1)y = 0, \qquad y' := \frac{dy}{dx}$ 2.1.1

For this problem, we take the specific case n = 1, in which the Legendre Differential Equation takes the form

 $\displaystyle L_2(y) := (1-x^2)y'' - 2xy' + 2y = 0$ 2.1.2

It is asserted that the following functions satisfy 2.1.2:

 $\displaystyle y_h^1(x) = x$ 2.1.3 $\displaystyle y_h^2(x) = \frac{x}{2}\log\left(\frac{1+x}{1-x}\right)-1$ 2.1.4

## Find

Show that $y_h^1(x)$ and $y_h^2(x)$ satisfy 2.1.2.

## Solution

This problem was solved without referring to past solutions.

$y_h^1$ is by far the simpler of the two functions, so we prove that it satisfies 2.1.2 first in order to make the process clear.
First, we find the first and second derivatives of $\displaystyle y_h^1$:

 $\displaystyle y_h^1 = x$ $\displaystyle {y_h^1}' = 1$ 2.1.5 $\displaystyle {y_h^1}'' = 0$ 2.1.6

Substituting 2.1.5 and 2.1.6 into 2.1.2 yields

 \displaystyle \begin{align} &(1-x^2)\cdot (0) - 2x\cdot (1) + 2\cdot (x) \\&= 0 - 2x + 2x \\&= 0 \end{align}

which proves that $y_h^1$ satisfies equation 2.1.2.

The process for showing that $y_h^2$ satisfies 2.1.2 is exactly the same as above.
We begin by carefully taking the first two derivatives of $y_h^2$:

$\displaystyle y_h^2(x) = \frac{x}{2}\log\left(\frac{1+x}{1-x}\right)-1$

 \displaystyle \begin{align} {y_h^2}' &= \frac{d}{dx}\left(\frac{x}{2}\right)\cdot\log\left(\frac{1+x}{1-x}\right) + \frac{x}{2}\cdot\frac{d}{dx}\left(\log\left(\frac{1+x}{1-x}\right)\right) \\&= \frac{1}{2}\cdot\log\left(\frac{1+x}{1-x}\right) + \frac{x}{2}\cdot\left(\frac{1-x}{1+x}\right)\cdot\frac{d}{dx}\left(\frac{1+x}{1-x}\right) \\&= \frac{1}{2}\cdot\log\left(\frac{1+x}{1-x}\right) + \frac{x}{2}\cdot\left(\frac{1-x}{1+x}\right)\cdot\left(\frac{1}{1-x} + \frac{1+x}{{\left(1-x\right)}^2}\right) \end{align}

Simplifying the last term yields

 $\displaystyle {y_h^2}' = \frac{1}{2}\cdot\log\left(\frac{1+x}{1-x}\right) - \frac{x}{x^2-1}$ 2.1.7

For the second derivative, we can easily differentiate the first term by using the results from taking the first derivative.

 \displaystyle \begin{align} {y_h^2}'' &= \frac{d}{dx}\left(\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)\right) - \frac{d}{dx}\left(\frac{x}{x^2-1}\right) \\&= -\frac{1}{x^2-1} - \left(\frac{1}{x^2-1} - \frac{2x^2}{{\left(x^2-1\right)}^2}\right) \end{align}

simplifying yields

 $\displaystyle {y_h^2}'' = \frac{2}{{\left(x^2-1\right)}^2}$ 2.1.8

Finally, substituting 2.1.7 and 2.1.8 into 2.1.2 yields

 \displaystyle \begin{align} &\left(1-x^2\right)\cdot\left(\frac{2}{{\left(x^2-1\right)}^2}\right) - 2x\cdot\left(\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)-\frac{x}{x^2-1}\right) + 2\cdot\left(\frac{x}{2}\log\left(\frac{1+x}{1-x}\right)-1\right) \\&=-\frac{2}{x^2-1} - x\log\left(\frac{1+x}{1-x}\right) + \frac{2x^2}{x^2-1} + x\log\left(\frac{1+x}{1-x}\right) - 2 \\&=\frac{2x^2-2}{x^2-1} - 2 \\&= 0 \end{align}

which proves that $y_h^2$ satisfies 2.1.2.

Since the Legendre Differential Equation is second-order, and we now possess two linearly independent solutions to the homogenous equation (see R2.6), every solution of the homogenous form of the equation can be expressed as a linear combination of $y_h^1$ and $y_h^2$.

# R*2.2- Reduced-Order Solution of an ODE with Missing Dependent Variable

## Given

$\displaystyle y''+ y' = x$

$\displaystyle p(x):=y'(x)$

$\displaystyle p' + p = x$

(2.2.1)

Solution is that

$\displaystyle p(x)= k_1 e^{-x}+x-1$

(2.2.2)

## Find

Verify that eq (2.2.2) is indeed solution for eq (2.2.1).

## Solution

Solved on my own

Equation 2.2.1 is that

$\displaystyle p' + p = x$

(2.2.1)

And now Let $\displaystyle T = x - p$

$\displaystyle \frac {dT}{dx}= 1 - p'$

$\displaystyle p'=1- \frac {dT}{dx}$

Plug in eq (2.2.1)

The equation 2.2.1 is following that

$\displaystyle 1- \frac{dT}{dx} = t$

$\displaystyle 1 - T = \frac {dT}{dx}$

$\displaystyle dx = -\frac {dT}{(T-1)}$

$\displaystyle -k - x = ln(T-1)$

$\displaystyle T - 1 = e^{-k-x}$

$\displaystyle T= 1 - ke^{-x}$

$\displaystyle p(x)= x - T = ke^{-x} + x - 1$

(2.2.2)

It shows that eq (2.2.2) is indeed solution for eq (2.2.1).

If you want double check,

$\displaystyle p=ke^{-x} + x - 1$

$\displaystyle p'= -ke^{x}+1$

Thus,

$\displaystyle p'+p=ke^{-x} + x - 1-ke^{x}+1=x$

# R*2.3- Linearity and Nonlinearity in General First-Order ODEs

Solved on my own

## Given

An ODE of 1st order is read as

$\displaystyle M(x,y)+N(x,y)\,y'=0$

(2.3.1)

## Find

show that the eqn (2.3.1) is linear in $\displaystyle y'$, and that it is in general an N1-ODE. But it is not the most general N1-ODE as represented by

$\displaystyle G(y',y,x)=0$

(2.3.2)

Give an example of a more general N1-ODE.

## Solution

#### Definition of Linear

If $\displaystyle L_{1}$ is a linear operator, then it must satisfy $\displaystyle L_{1}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)$

Where $\displaystyle u,v$ are the independent variables, and $\displaystyle \alpha, \beta \in \mathbb{R}.$

$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=-[\alpha(\frac{M}{N})_{1}+\beta(\frac{M}{N})_{2}]$

$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=-\alpha(\frac{M}{N})_{1}-\beta(\frac{M}{N})_{2}$

$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=\alpha L_{1}(y'i_{1})+\beta L_{1}(y'_{2})$

$\displaystyle \therefore M(x,y)\,dx+N(x,y)\,y'=0$ is linear in $\displaystyle y'$.

And it is easy to proof that

$\displaystyle L_{1}(\alpha y_{1}+\beta y_{2})\neq\alpha L_{1}(y_1)+\beta L_{1}(y_2)$

So, is in general an N1-ODE.

## An example of the general N1-ODE

$\displaystyle M(x,y)+N(x,y)f(y')=0$.

(2.3.3)

where $\displaystyle f(y')$ is an arbitrary function of $\displaystyle y'$.

# R*2.4- Verification of Nonlinearity in a Particular First-Order ODE

## Given

 $\displaystyle (4x^7+siny)+(x^2y^3)\,y'=0$ (2.4.1)

## Find

Verify that (2.4.1) is a N1-ODE.

## Solution

Solved on my own
 $\displaystyle G\left(y',y,x\right)=0$ (2.4.2)

(2.4.1) is a statement of (2.4.2) showing that the highest order is one. Thus, (2.4.1) is a fist order differential equation.

Linearty has to satisfy the following condition:

 $\displaystyle F(\alpha u+\beta v)=\alpha F(u)+\beta F(v), \forall \alpha, \beta\in\mathbb R$ (2.4.3)

Nonlinearty is the condition not satisfing the above statement as follows:

 $\displaystyle F(\alpha u+\beta v)\neq\alpha F(u)+\beta F(v)$ (2.4.4)

(2.4.1) can be rewritten as follows:

 $\displaystyle F(y)=(4x^7+siny)+(x^2y^3)\,y'$ (2.4.5)

And then, the nonlinearity can be checked with the following two functions:

 $\displaystyle F(\alpha y)=(4x^7+sin(\alpha y))+(x^2(\alpha y)^3)\,y'$ (2.4.6)
 $\displaystyle \alpha F(y)=\alpha(4x^7+siny)+\alpha(x^2y^3)\,y'$ (2.4.7)

From (2.4.6) and (2.4.7), $\displaystyle F(\alpha y)\neq\alpha F(y)$ is found so that (2.4.1) is verified to be a nonlinear equation.

Finally it is concluded that (2.4.1) is a N1-ODE (Nonlinear First Oder Differential Equation).

# R*2.5- Converting a General First-Order ODE into Particular Form

## Given

 $\displaystyle \bar M(x,y)+ \bar N(x,y)(y')^3=0$ 2.5.1

Equation 2.5.1 satisfies the most general form $\displaystyle G\left(y',y,x \right)= 0$, but can be converted into particular form $\displaystyle M(x,y)+N(x,y)\,y'=0$.

## Find

Using the hint:

 $\displaystyle M(x,y):=\left[\bar M(x,y)\right]^{1/3}$ 2.5.2

and

 $\displaystyle N(x,y):=\left[\bar N(x,y)\right]^{1/3}$ 2.5.3

Illustrate a way to convert equation 2.5.1 into particular form $\displaystyle M(x,y)+N(x,y)\,y'=0$.

## Solution

Solved on my own

$\displaystyle \bar M(x,y)+ \bar N(x,y)(y')^3=0$

$\displaystyle \bar N(x,y)(y')^3=-\bar M(x,y)$

$\displaystyle y'=\left[-\frac{\bar M(x,y)}{\bar N(x,y)}\right]^{1/3}$

 $\displaystyle \left[\bar M(x,y)\right]^{1/3}+\left[\bar N(x,y)\right]^{1/3}y'=0$ 2.5.4

Let
$\displaystyle M(x,y):=\left[\bar M(x,y)\right]^{1/3}$
And
$\displaystyle N(x,y):=\left[\bar N(x,y)\right]^{1/3}$

Equation 2.5.4 becomes $\displaystyle M(x,y)+N(x,y)\,y'=0$ which is a particular form of Equation 2.5.1 .

# R*2.6 - Linear Independence of Solutions to the Second-Order Legendre ODE

## Given

The Legendre Differential Operator/Equation takes the form:

$\displaystyle L_2(y) := (1-x^2)y''-2xy'+n(n+1)y=0$ (1)

$\displaystyle n=1 \Rightarrow (1-x^2)y''-2xy'+2y=0$ (2)

Two homogeneous solutions for this equation are:

$\displaystyle y^1_{H}(x) = x$ (3)

$\displaystyle y^2_{H}(x) = \frac{x}{2} \log \left(\frac{1+x}{1-x} \right) - 1$ (4)

## Find

Show that equations (3) and (4) are linearly independent, i.e., show that

$\forall \alpha \in \mathbb R, y^1_{H}(\cdot) \ne \alpha \, y^2_{H}(\cdot)$

i.e., for any given $\alpha$, show that

$\exists \hat x \rm \ such \ that \ y^1_{H}(\hat x) \ne \alpha y^2_{H}(\hat x)$

Plot $y^1_{H}(x)$ and $y^2_{H}(x)$.

## Solution

Solved on my own

Since we know there exists an $\alpha$ and only one $\alpha$ such that for $\hat x_1$

$y^1_{H}(\hat x_1)=\alpha_1 y^2_{H}(\hat x_1)$

We also know that for any other point $\hat x_2$ there exists an $\alpha_2$ such that

$y^1_{H}(\hat x_2)=\alpha_2 y^2_{H}(\hat x_2)$

The two homogeneous solutions are linearly independent if $\alpha_1 \ne \alpha_2$
For $\hat x_1 = \frac{1}{2}$

$\displaystyle \frac{1}{2} = \alpha_1 \left[\frac{\frac{1}{2}}{2} \log \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right) - 1 \right]$

Solving for $\alpha_1$ we determine $\alpha_1$ = -0.5677175.

For $\hat x_2 = \frac{1}{4}$

$\displaystyle \frac{1}{4} = \alpha_2 \left[\frac{\frac{1}{4}}{2} \log \left(\frac{1+\frac{1}{4}}{1-\frac{1}{4}} \right) - 1 \right]$

Solving for $\alpha_2$ we determine $\alpha_2$ = -0.2571305.

Since $\alpha_1 \ne \alpha_2$ the homogeneous solutions must be linearly independent. This can be seen by looking at their graph.

File:HomogeneousSolution.png
The two homogeneous solutions for the Legendre Differential Operator/Equation for n = 1

# R*2.7 - Deriving Exact First Order Differential Equations from Level Sets

## Given

The family of level sets

 $\displaystyle \phi(x,y) = x^2y^{\frac{3}{2}} + \log(x^3y^2) = k$ 2.7.1

## Find

Find

$G(y',y,x) = \frac{d}{dx}\phi(x,y) = 0$

and show that it is an N1-ODE.

## Solution

This solution was found without referring to past solutions.

We know from the chain-rule[2] that

$\frac{d\phi}{dx} = \frac{\partial\phi}{\partial x} + \frac{\partial\phi}{\partial y}\frac{\partial y}{\partial x}$

Applying this directly yields

\begin{align} \frac{\partial\phi}{\partial x} &= 2xy^{\frac{3}{2}} + 3x^{-1} \\ \frac{\partial\phi}{\partial y} &= \frac{3}{2}x^2y^{\frac{1}{2}} + 2y^{-1} \end{align}

Thus

 $\displaystyle G(y',y,x) = \underbrace{2xy^{\frac{3}{2}} + 3x^{-1}}_M + \underbrace{\left(\frac{3}{2}x^2y^{\frac{1}{2}} + 2y^{-1}\right)}_Ny' = 0, \qquad y' := \frac{\partial y}{\partial x}$ 2.7.2

which, as can be seen, is already in the form $M(x,y) + N(x,y)y' = 0$.
Due to the way that it was derived, it is expected that 2.7.2 is exact, and indeed

$\displaystyle M_{,y} = 3xy^{\frac{1}{2}} = N_{,x} \qquad M_{,y} := \frac{\partial M}{\partial y}$

It is obvious that G is an ordinary differential equation and that it is first-order.
To prove that it is nonlinear, suppose that $y_1$ and $y_2$ are functions of x. Then, slightly changing our notation to G(y) for clarity, we see that

\begin{align} \displaystyle G(\alpha\cdot y_1 + \beta\cdot y_2) &= 2x\left(\alpha\cdot y_1 + \beta\cdot y_2\right)^{\frac{3}{2}} + 3x^{-1} \\ &\qquad + \left(\frac{3}{2}x^2\left(\alpha\cdot y_1 + \beta\cdot y_2\right)^{\frac{1}{2}} + 2\left(\alpha\cdot y_1 + \beta\cdot y_2\right)^{-1}\right)\left(\alpha\cdot y_1 + \beta\cdot y_2\right)' \\ &\neq \alpha\cdot G(y_1) + \beta\cdot G(y_2) \end{align}

The powers of y in G make it nonlinear.

# R*2.8 - First Exactness Condition

## Given

First exactness condition is that an ODE can be written in the form

$\displaystyle M(x,y) + N(x,y)\frac {dy}{dx} = 0$

## Find

Does the following N1-ODE satisfy the first exactness condition?

$\displaystyle M(x,y)cosy' + N(x,y)logy' = 0$

(2.8.1)

## Solution

Solved on my own

In order to satisfy the first exactness condition, it must be in the form of:

$\displaystyle M(x,y) + N(x,y)\frac {dy}{dx} = 0$

$\displaystyle y'= -\frac{M(x,y)}{N(x,y)}$

(2.8.2)

So rearrange eq (2.8.1) to fit eq (2.8.2)

$\displaystyle M(x,y)cosy' + N(x,y)logy' = 0$

$\displaystyle M(x,y) + N(x,y)\frac{logy'}{cosy'}=0$

$\displaystyle f(y')=\frac{logy'}{cosy'}$

$\displaystyle f(y')$ is any function of y'. If the function $\displaystyle f(y')$ has no explicit inverse $\displaystyle f^{-1}(y'),$

then it can not satisfy in the form of eq (2.8.2)

In this case, $\displaystyle f(y')$ does not have exact inverse $\displaystyle f^{-1}(y')$

Thus, This does not satisfy the first exactness condition.

# R*2.9- Commutative Partial Derivatives

Solved on my own

## Given

Review calculus, and differentiation

$\displaystyle \frac{\partial^{2}\phi(x,y)}{\partial x\partial y}=\frac{\partial^{2}\phi(x,y)}{\partial y\partial x}$

(2.9.1)

## Find

Find the minimum degree of differentiability of the function $\displaystyle \phi(x,y)$ such that the above equation is satisfied, State the full theorem and provide a proof.

## Solution

#### Clairaut's theorem

According to the Schwarz's theorem,[3], if

$\displaystyle f \colon \mathbb{R}^n \to \mathbb{R}$

has continuous second partial derivatives at any given point in $\displaystyle \mathbb{R}^n$, then for $\displaystyle 1 \leq i,j \leq n,$

$\displaystyle \frac{\partial^2 f}{\partial x_i\, \partial x_j}(a_1, \dots, a_n) = \frac{\partial^2 f}{\partial x_j\, \partial x_i}(a_1, \dots, a_n).\,\!$

And we then know that the minimum degree of differentiability of the function $\displaystyle \phi(x,y)$ must be three when it satisfies eqn (2.9.1)

## Proof of Clairaut's theorem

The result was first discovered by L.Euler around 1734 in connection with problems in hydrodynamics [4]. In order to prove this theorem, we shall first introduce the Mean Value theorem as follows:

Let $\displaystyle f:[a,b]\rightarrow\mathbb{R}$ be a continuous function on the closed interval $\displaystyle [a,b]$, and derivative on the open interval $\displaystyle (a,b)$, where $\displaystyle a, Then $\displaystyle \exists c\in[a,b]$ such that

$f ' (c) = \frac{f(b) - f(a)}{b - a}.$

(2.9.2)

Suppose Q is a closed rectangle with sides parallel to the coordinate axes, put

$\triangle(f,Q)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)$

(2.9.3)

$\ u(t)=f(t,b+k)-f(t,b)$

$\triangle(f,Q)=u(a+h)-u(a)$

$\ =hu'(x)$

$=h(\frac{\partial f}{\partial x}(x,b+k)-\frac{\partial f}{\partial x}(x,b)$

$=hk\frac{\partial^{2}}{\partial y\partial x}f(x,y)$

(2.9.4)

So, we put

$A=\frac{\partial^{2}}{\partial y\partial x}f(a,b)$,

Chose $\displaystyle \epsilon>0$. if $\displaystyle h$ and $\displaystyle k$ are sufficiently small, we have

$|\frac{\triangle(f,Q)}{hk}-A|<\epsilon$

(2.9.5)

Fix $\displaystyle h$, and let $\displaystyle k\rightarrow0$, then

$|\frac{(\frac{\partial f}{\partial y}(a+h,b)-\frac{\partial f}{\partial y}(a,b)}{h}-A|\leq\epsilon$

(2.9.6)

Since $\displaystyle \epsilon$ was is arbitrary, and (2.9.6) holds for all sufficiently small $\displaystyle h\neq0$

$\frac{\partial^{2}}{\partial y\partial x}f(a,b)=\frac{\partial^{2}}{\partial x\partial y}f(a,b)$

(2.9.7)

Since $\displaystyle (a,b)$ is an arbitrary point in $\displaystyle \mathbb{R}^2$, so

$\frac{\partial^{2}}{\partial y\partial x}f(x,y)=\frac{\partial^{2}}{\partial x\partial y}f(x,y)$

# R*2.10-Verification of solution for N1-ODE

## Given

 $\displaystyle y(x)=\sin^{-1}(k-15x^5)$ (2.10.1)
 $\displaystyle M(x,y)+N(x,y)\,y'=0$ (2.10.2)

where $\displaystyle M(x,y)=75x^4$ and $\displaystyle N(x,y)=cos y$

## Find

Verify that (2.10.1) is the solution for the (2.10.2).

## Solution

Solved on my own

(2.10.1) would be rewritten as follows:

 $\displaystyle \sin y=k-15x^5$ (2.10.3)

Differentiating the both sides of (2.10.3), (2.10.3) results in

 $\displaystyle (\cos y)y'=-75x^4 \Rightarrow 75x^4+(\cos y)y'=0$ (2.10.4)

Thus, (2.10.1) is verified to be the solution for (2.10.2.)

# R 2.11-Solving PDE for Integrating Factor

## Given

For

 $\displaystyle h_x N - h_y M + h(N_x - M_y)=0$ 2.11.1

## Find

Why solving equation 2.11.1 for the integrating factor $\displaystyle h(x,y)$ is usually not easy ?

## Solution

Solving equation 2.11.1 for the integrating factor $\displaystyle h(x,y)$ is usually not easy because in general equation 2.11.1 is a non-linear partial differential equation for unknown function $\displaystyle h$ and it is having terms $\displaystyle h_x, h_y, N_x, M_y$ which makes it very difficult to solve for $\displaystyle h(x,y)$.

# R*2.12 - Case 2: Finding $h$ as a Function of y

## Given

A N1-ODE satisfies the first exactness condition if it takes the particular form:

$\displaystyle M(x,y) + N(x,y)y'$ (1)

The second exactness condition states that

$\displaystyle M_{y}(x,y) = N_{x}(x,y) \ i.e., \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x}$ (2)

If a N1-ODE satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor $h(x,y)$ be found such that the following N1-ODE is exact:

$\displaystyle h(x,y) \left[M(x,y)+N(x,y) \, y' \right]=0$ (3)

This can be written as:

$\displaystyle \underbrace{(hM)}_{\displaystyle\color{blue}{\bar M}} + \underbrace{(hN)}_{\displaystyle\color{blue}{\bar N}} \, y' = 0$ (4)

To find $h$ we must apply the second exactness condition to obtain:

$\displaystyle h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$ (5)

## Find

Suppose $h_{x}(x,y)=0$, thus $h$ is a function of y only, then equation (5) becomes:

$\displaystyle \frac{h_{y}}{h}=\frac{1}{M}(N_{x}-M{y})=: m(y)$ (6)

Find $h$ using equation (6).

## Solution

Solved on my own

To find $h$ we can integrate equation (6) to obtain:

$\displaystyle \int \frac{h_{y}}{h}= \int \frac{1}{M}(N_{x}-M{y})$ (7)

$\displaystyle \ln(h(y)) = \int^y m(s)ds + k$

$\displaystyle h(y) = exp \left[\int^y m(s)ds + k \right]$

# R*2.13 - Using an Integrating Factor to Render an Inexact L1-ODE-VC Exact

## Given

 $\displaystyle y' + x^{-1}y = x^2$ 2.13.1

## Find

Show that h(x) = x is an integrating factor for 2.13.1, and that

$y(x) = \frac{x^3}{4} + \frac{k}{x}$

is its solution.

## Solution

This solution was found without reference to past solutions.

We first show that 2.13.1 is not already exact, or else its integrating factor would be $1$!

Re-arranging 2.13.1, we arrive at the standard form

$\displaystyle \underbrace{x^{-1}y-x^2}_M + \underbrace{1}_N y' = 0$

Now,

$\displaystyle M_{,y} = x^{-1} \quad\neq\quad 0 = N_{,x}, \qquad M_{,y} := \frac{\partial M}{\partial y}$

so 2.13.1 is not exact.

We thus proceed to multiply 2.13.2 by h(x) = x which yields

 $\displaystyle \underbrace{y - x^3}_{\bar{M}} + \underbrace{x}_{\bar{N}} y' = 0$ 2.13.2

Now,

$\displaystyle \bar{M}_{,y} = 1 = \bar{N}_{,x}$

so 2.13.2 is exact and h(x) = x is the integrating factor that makes 2.13.1 exact.

Since every exact differential equation may be produced from a first integral $\phi(x,y) = k$ such that

$M(x,y) = \frac{\partial\phi}{\partial x}$

and

$N(x,y) = \frac{\partial\phi}{\partial y}$

we may integrate $\bar{M}$ and $\bar{N}$ to find $\phi$ as follows:

 $\displaystyle \int{\bar{M}(x,y)}\partial x = \int{(y - x^3)}\partial x = xy - \frac{x^4}{4} + \psi(y)$ 2.13.3
 $\displaystyle \int{\bar{N}(x,y)}\partial y = \int{x}\partial y = xy + \theta(x)$ 2.13.4

Combining 2.13.3 and 2.13.4 yields

$\displaystyle \phi(x,y) = k = xy - \frac{x^4}{4}$

which upon solving for $y$ yields

$\displaystyle y = \frac{x^3}{4} + \frac{k}{x}$

While this already proves the assertion, we can also show that this is the correct expression for $y$ by plugging it back in to equation 2.13.1:

\displaystyle \begin{align} y' &= \frac{3}{4}x^2 - \frac{k}{x^2} \\y' + x^{-1}y &= \left(\frac{3}{4}x^2 - \frac{k}{x^2}\right) + x^{-1}\left(\frac{x^3}{4} + \frac{k}{x}\right) \\ &= x^2 \end{align}

# R*2.14 - General L1-ODE-VC

## Given

$\displaystyle h(x)=exp\left[\int^x a_0(s)ds + k_1\right]$

(2.14.1)

$\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$

(2.14.2)

$\displaystyle a_1(x)y'+a_0(x)y= b(x)$

(2.14.3)

## Find

Solve the general L1-ODE-VC

1. $\displaystyle a_1(x)=1$

$\displaystyle a_0(x)=x$

$\displaystyle b(x)=2x + 3$

2. Find $\displaystyle y(x)$ in term of $\displaystyle a_1(x), a_0(x), b(x)$

3. $\displaystyle a_1(x)=x^2+1$

$\displaystyle a_0(x)=x$

$\displaystyle b(x)=2x$

## Solution

Solved on my own

an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus, in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential.

From lecture note,

$\underbrace{\left( hM \right)}_{{\bar{M}}}+\underbrace{\left( hN \right)}_{{\bar{N}}}{y}'=0$

And applying second exactness condition, the equation is following that

$\underbrace{\left( h\frac{\partial M}{\partial y}+M\frac{\partial h}{\partial y} \right)}_{\frac{\partial \bar{M}}{\partial y}}=\underbrace{\left( h\frac{\partial N}{\partial x}+N\frac{\partial h}{\partial x} \right)}_{\frac{\partial \bar{N}}{\partial x}}$

If we assume that $\displaystyle h_y(x,y)=0$, then h is function of x only and the equation is following that

$\left( h\frac{\partial M}{\partial y} \right)=\left( h\frac{\partial N}{\partial x}+N\frac{\partial h}{\partial x} \right)$

And we obtain some equations through rearranging equation above

$\displaystyle h{{N}_{x}}+N{{h}_{x}}-h{{M}_{y}}=0$

$\displaystyle {{N}_{x}}=\frac{\partial N}{\partial x}$

$\displaystyle {{M}_{y}}=\frac{\partial M}{\partial y}$

$\displaystyle {{h}_{x}}=\frac{\partial h}{\partial x}$

And we can obtain h(x) through dividing by h

$\displaystyle \frac {h_x}{h}=-\frac{1}{N}(\underbrace{N_x}_{0}-\underbrace{M_y}_{a_0(x)})=a_0(x)$

$\displaystyle h(x)={{e}^{-\int{\frac{1}{N}\left( {{N}_{x}}-{{M}_{y}} \right)dx}}}=exp\left[\int^x a_0(s)ds + k_1\right]$

(2.14.1)

And in order to obtain y(x), we can apply the integrating factor to obtain the following equations

$\displaystyle y'+a_0y=b \rightarrow h(y'+a_0y)=hb$

and $\displaystyle h_x/h=a_0 \rightarrow ha_0=h_x=h'$

$\displaystyle hy'+h'y=hb$

Thus, $\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$

(2.14.2)

### part 1

1. In order to solve this problems, plug given information in eq (2.14.3)

the equation is following that

$\displaystyle y'+ xy = 2x + 3$

Also, eq (2.14.1) is following that

$\displaystyle h(x)=exp\left[\int^x sds+k_1\right]$

$\displaystyle h(x)=exp\left[\frac {1}{2}s^2\right]$

And eq (2.14.2) is following that

$\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$

$\displaystyle = \frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x exp\left[\frac {1}{2}s^2\right]*(2s+3)ds+k_2\right]$

$\displaystyle =\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x e^{\frac {1}{2}s^2}*2sds+\int^x e^{\frac {1}{2}s^2}*3ds\right]$

Let $\displaystyle t= \frac {s^2}{2}$

$\displaystyle \frac {dt}{s}=ds$

Finally, the equation is that

$\displaystyle y(x) =\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x 2e^tdt+\int^x e^{\frac {1}{2}s^2}*3ds\right]=\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[2e^{\frac {x^2}{2}}+\int^x e^{\frac {1}{2}s^2}*3ds\right]$

In order to solve more, numerical method must be used.

### part 2

From eq 2.14.3

$\displaystyle a_1(x)y'+a_0(x)y= b(x)$

$\displaystyle y'+ \frac {a_0(x)}{a_1(x)}y=\frac {b(x)}{a_1(x)}$

and plug in eq 2.14.1

$\displaystyle h(x)=exp\left[\int^x \frac {a_0(s)}{a_1(s)}ds\right]$

$\displaystyle y(x)=\frac {1}{exp\left[\int^x \frac {a_0(s)}{a_1(s)}ds\right]}\left[\int^x exp \left[ \int^x \frac {a_0(s)}{a_1(s)}ds \right] \frac{b(s)}{a_1(s)}ds\right]$

### part 3

$\displaystyle (x^2+1)y'+xy=2x$

$\displaystyle y'+\frac {x}{x^2+1}y=\frac {2x}{x^2+1}$

At this point,

$\displaystyle h(x)=exp\left[ \int^x \frac {s}{s^2+1}ds \right]$

Let $\displaystyle t=s^2+1$

$\displaystyle ds = \frac {dt}{2s}$

$\displaystyle h(x)=exp \left[ \int^x \frac {dt}{2t} \right]$

$\displaystyle h(x)=exp \left[ \frac {1}{2}log(x^2+1) \right] = \sqrt{{x^2}+{1}}$

(2.14.4)

Then plug in eq 2.14.2

$\displaystyle y(x)=\frac {1}{\sqrt{{x^2}+{1}}}\left[ \int^x \frac {2s}{\sqrt{{s^2}+{1}}}ds\right]$

Let $\displaystyle u= s^2 + 1$

$\displaystyle du = 2sds \rightarrow ds= \frac {du}{2s}$

$\displaystyle y(x)=\frac {1}{\sqrt{{x^2}+{1}}}\left[ \int^x u^{-\frac {1}{2}}du \right] = \frac {1}{\sqrt{{x^2}+{1}}}\left[ 2 \sqrt{{x^2}+{1}}+ C \right]$

Final solution is that

$\displaystyle y(x)= 2 + \frac {C}{\sqrt{{x^2}+{1}}}$

# R*2.15- Integration constant

Solved on my own

## Given

Since

$\displaystyle P(x)y'+Q(x)y=R(x)$

(2.15.1)

$\displaystyle y'+\frac {Q(x)}{P(x)}y=\frac{R(x)}{P(x)}$

(2.15.2)

is an L1-ODE_VC, there should be only one integration constant, not two.

## Find

Show that the integration constant $\displaystyle k_{1}$ in

$\displaystyle h(x)=exp[\int^{x}a_{0}(s)ds+k_{1}]$

(2.15.3)

is not necessary, i.e., only $\displaystyle k_{2}$ in

$\displaystyle y(x)=\frac{1}{h(x)}[\int^{x}h(s)b(s)ds+k_{2}]$

(2.15.4)

is necessary.

## Solution

Substituting the eqn (2.15.3) to eqn (2.15.4), we obtain the following expression

$\displaystyle y(x)=\frac{1}{e^{k_{1}}\cdot e^{\int^{x}a_{0}(s)ds}}[e^{k_{1}}\cdot\int^{x}e^{\int^{x}a_{0}(s)ds}b(s)ds+k_{2}]$

$\displaystyle y(x)=\frac{1}{e^{\int^{x}a_{0}(s)ds}}[\int^{x}e^{\int^{x}a_{0}(s)ds}b(s)ds+\frac{k_{2}}{e^{k_{1}}}]$

Then, we can define a new constant $\displaystyle \bar{k}_{2}$ such that

$\displaystyle \bar{k}_{2}=\frac{k_{2}}{e^{k_{1}}}$

hence, this new constant contains $\displaystyle k_{1}$. In other words, $\displaystyle k_{1}$ is unnecessary.

# R*2.16-Solution of L1-ODE-VC

## Given

 $\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k_2\right]$ (2.16.1)

 $\displaystyle h(x)=\exp\left[{\int^x a_0(s)ds+k_1}\right]$ (2.16.2)

The derivated result in King 2003 p.512 is

 $\displaystyle y(x)=A\,y_H(x)+y_P(x)$ (2.16.3)

## Find

Show that (2.16.1) agrees with (2.16.3).

## Solution

Solved on my own

 $\displaystyle a_0(x)y'+a_0(x)y=b(x)$ (2.16.4)

In King 2003 p.512, the L1-ODE-VC is presented as follows:

 $\displaystyle y'+P(x)y=Q(x)$ (2.16.5)

From (2.16.3) and (2.164), the following relations can be derived.

 $\displaystyle a_1(x)=1$ (2.16.6)
 $\displaystyle a_0(x)=P(x)$ (2.16.7)
 $\displaystyle b(x)=Q(x)$ (2.16.8)

From the relations, in King 2003 P512, the integrating factor is shwon as follows:

 $\displaystyle h(x)=\exp{\int^x P(t)dt}$ (2.16.9)

When (2.16.1) is substituted by (2.16.9), (2.16.1) becomes

 $\displaystyle y(x) = \exp\left(-\int^x P(t)dt\right)\left[{\int^x \exp\left(\int^x P(t)dt\right)b(s)ds+k_2}\right]$ (2.16.10)

$\displaystyle \Rightarrow y(x)= k_2\exp\left(-\int^x P(t)dt\right)+\exp\left(-\int^x P(t)dt\right)\int^x \exp\left(\int^x P(t)dt\right)b(s)ds$

Substituting the relation (2.16.8) for (2.16.10), (2.16.10) can be rewritten as follows:

 $\displaystyle y(x)= k_2\exp\left(-\int^x P(t)dt\right)+\exp\left(-\int^x P(t)dt\right)\int^x \exp\left(\int^x P(t)dt\right)Q(s)ds$ (2.16.11)

(2.16.10) is the solution presented in King 2003 p.512 identifying A, $\displaystyle y_H(x)$ and $\displaystyle y_p(x)$ as follows:

 $\displaystyle A=k_2$ (2.16.12)
 $\displaystyle y_H(x)=exp\left(-\int^x P(t)dt\right)$ (2.16.13)
 $\displaystyle y_P(x)=\exp\left(-\int^x P(t)dt\right)\int^x \exp\left(\int^x P(t)dt\right)Q(s)ds$ (2.16.14)

Thus, it is verified that (2.16.1) agrees with (2.16.3).

# R*2.17-Solving for homogenous counterpart

## Given

Instead of identifying $\displaystyle y_H(x)$ from $\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds + k_2\right]$ .

## Find

Solve the homogenous counterpart of $\displaystyle y' + a_0(x)\,y=0$ .

## Solution

Solved on my own

$\displaystyle y' + a_0(x)\,y=0$

$\displaystyle y'= -a_0(x)y$

$\displaystyle \frac{dy}{y}=-a_0(x)dx$

$\displaystyle \int^y \frac{dy}{y}=-\int^s a_0(s)ds$

$\displaystyle y(x)=exp \left[-\int^x a_0(s)ds + k \right]$

$\Rightarrow y_H(x)=exp \left[-\int^x a_0(s)ds + k \right]$

# R*2.18 - Integrating Factor Method

## Given

A L1-ODE-VC satisfies the first exactness condition if it takes the particular form:

$\displaystyle M(x,y) + N(x,y)y'$ (1)

The second exactness condition states that

$\displaystyle M_{y}(x,y) = N_{x}(x,y) \ i.e., \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x}$ (2)

If a L1-ODE-VC satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor $h(x,y)$ be found such that the following N1-ODE is exact:

$\displaystyle h(x,y) \left[M(x,y)+N(x,y) \, y' \right]=0$ (3)

## Find

$\displaystyle \left [x^4 y + 10 \right ] + (\frac{1}{2} x^2)y'=0$ (4)

$\displaystyle M(x,y) = x^4 y + 10$

$\displaystyle N(x) = \frac{1}{2} x^2$

Find out whether equation (4) is exact. If it is not exact find the integrating factor $h$ to make it exact.

## Solution

Solved on my own

Since equation (4) satisfies the first exactness condition we must test the second exactness condition.

$\displaystyle M_{y} = x^4$

$\displaystyle N_{x} = x$

$\displaystyle x^4 \ne x$

Therefore, equation (4) is not exact. To make it exact an integration factor $h$ must be found such that:

$\displaystyle \frac{h_{x}}{h}= -\frac{1}{N(x)}(N_{x}-M{y})=: n(x)$

Solving for h(x) produces:

$\displaystyle h(x) = exp \left[\int^x n(s)ds + k \right]$

$\displaystyle h(x) = exp \left[2 \int^x (s^2 - \frac{1}{s})ds + k \right]$

$\displaystyle h(x) = exp \left[2 (\frac{1}{3} x^3 - \ln x) + k \right]$

# Contributing Members

 Team Contribution Table Problem Number Assigned To Solved By Typed By Proofread By Signature 2.1 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 2.2 kim kim kim all Egm6321.f11.team2.kim 19:51, 21 September 2011 (UTC) 2.3 Langshah Langshah Langshah all 2.4 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 2.5 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 2.6 Gary Gary Gary all 2.7 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 2.8 Kim Kim Kim all Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC) 2.9 Langshah Langshah Langshah all 2.10 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 2.11 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 2.12 Gary Gary Gary all 2.13 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 2.14 Kim Kim Kim all Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC) 2.15 Langshah Langshah Langshah all 2.16 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 2.17 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 2.18 Gary Gary Gary all

Egm6321.f11.team2.rho 20:41, 21 September 2011 (UTC)

1. http://mathworld.wolfram.com/LegendreDifferentialEquation.html
2. http://en.wikipedia.org/wiki/Chain_rule