# Problem R3.1 - Show Eq. (1) p.13-2 is exact only if $k_1=d_1$ (constant) and Eq. (1) p.12-4 is a particular equation

From Mtg 13-3

From Mtg 11-2

From Mtg 12-4

## Given

Eq. (1) p.13-2 is $\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$.

Eq. (2) p.11-2 is $\displaystyle n(x):=(-1/N)(N_x-M_y)$.

Eq. (1) p.12-4 is $\displaystyle M+Ny'=[a(x)y+k_x(x)]+\bar b(x)y'=0$.

## Find

$\displaystyle a)$Show that the N1-ODE (1) p.13-2 satisfies the condition (2) p.11-2 that an integrating factor $\displaystyle h(x)$ can be found to render it exact, only if $\displaystyle k_1(y)=d_1$ (constant).

$\displaystyle b)$Show that (1) p.13-2 includes (1) p.12-4 as a particular case.

## Solution

Solved independently.

$\displaystyle a)$Eq. (1) p. 13-2 can be written as $\displaystyle N(x,y)y'+M(x,y)=0$ which yields the following:

$\displaystyle N(x,y)=\bar b(x,y)c(y)$

$\displaystyle M(x,y)=a(x)\bar c(x,y)$.

These two equations yield the following when differentiated:

$\displaystyle N_x=\bar b(x,y)c_x(y)+\bar b_x(x,y)c(y)$

$\displaystyle M_y=a(x)\bar c_y(x,y)+a_y(x)\bar c(x,y)$.

Now we plug these equations into Eq. (2) p. 11-2 to obtain

$\displaystyle n(x)=\frac{-1}{\bar b(x,y)c(y)}*[\bar b_x(x,y)c(y)-a(x)\bar c_y(x,y)]$

This simplifies to $\displaystyle n(x)=\frac{a(x)\bar c_y(x,y)}{\bar b(x,y)c(y)}-\frac{\bar b_x(x,y)c(y)}{\bar b(x,y)c(y)}$.

Since $\displaystyle \bar b_x(x,y)=b(x,y)$ & $\displaystyle \bar c_y(x,y)=c(x,y)$ cancel out, the formula reduces to

$\displaystyle n(x)=\frac{a(x)}{\bar b(x,y)}-1$.

Rearranging the equation produces $\displaystyle \bar b(x,y)=\frac{a(x)}{n(x)+1}$.

If you differentiate the above equation by $\displaystyle y$, you obtain

$\displaystyle \bar b_y(x,y)=0$ because everything is a function of $\displaystyle x$ or a constant.

This shows that $\displaystyle k_1(y)=d_1$ (constant) and not a function of $\displaystyle y$.

$\displaystyle b)$In order to show Eq. (1) p. 12-4 is a particular case of Eq. (1) p. 13-2, we must make two assumptions.

Those assumptions include $\displaystyle \bar b(x,y)=\bar b(x)$ and $\displaystyle c(y)=C$(constant).

Working first with $\displaystyle M(x,y)$,

we have $\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$

(left side is the general case and the right side is the particluar case).

With $\displaystyle c(y)=C$, Eq. (1) p. 13-3 reduces to $\displaystyle \bar c(x,y)=Cy+k_2(x)$.

Plug the above equation for $\displaystyle \bar c(x,y)$ into the $\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$ which yields

$\displaystyle a(x)[Cy+k_2(x)]=a(x)y+k_2(x)$.

The left side simplies becase $\displaystyle C$ is absorbed into $\displaystyle a(x)$ and $\displaystyle a(x)$ is absorbed into $\displaystyle k_2(x)$.

Working last with $\displaystyle N(x,y)$,

we have $\displaystyle \bar b(x,y)c(y)=\bar b(x)$

(left side is the general case and the right side is the particluar case).

Since $\displaystyle c(y)=C$, it can be absorbed into $\displaystyle \bar b(x,y)$.

Also, $\displaystyle \bar b(x,y)=\bar b(x)$ because it is not a function of $\displaystyle y$.

This ultimately yields the particluar case, $\displaystyle \bar b(x)$.

# Problem R*3.2 - Show Eq. (1) p.13-4 is exact or can be made exact by the IFM. Find the integrating factor h

From Mtg 13-4

## Given

Eq. (1) p.13-2 is $\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$ or $\displaystyle N(x,y)y'+M(x,y)=0$.

Eq. (2) p.13-2 is $\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y)$.

Eq. (1) p.13-3 is $\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x)$.

$\displaystyle a(x) = 5x^3+2$

$\displaystyle b(x) = x^2$

$\displaystyle c(x) = y^4$

$\displaystyle \bar b(x) = \frac{1}{3}x^3+d_1$

$\displaystyle \bar c(x) = \frac{1}{5}y^5+\sin(x)+d_2$

So, after combining the above equations into Eq. (1) p.13-4 we obtain

Eq. (1) p.13-4 which is $\displaystyle (\frac{1}{3}x^3+d_1)y^4y'+(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)=0$.

## Find

Show that Eq. (1) p. 13-4 either is exact or can be made exact by the IFM. Find the integrating factor $\displaystyle h$.

## Solution

Solved independently.

In order for Eq. (1) p. 13-4 to be exact, it must satisfy both conditions of exactness.

The first exactness condition requires the formula to be in the following form: $\displaystyle N(x,y)y'+M(x,y)=0$.

Eq. (1) p. 13-4 already satisfies the above requirement so it meets this condition.

The second exactness condition requires the formula to satisfy the following condition: $\displaystyle M_y(x,y)=N_x(x,y)$.

Differentiate the following equations, $\displaystyle M(x,y)=(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)$ and $\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4$,

to obtain $\displaystyle M_y(x,y)=5*(x^3+\frac{2}{5})y^4$ and $\displaystyle N_x(x,y)=x^2y^4$.

Clearly, $\displaystyle M_y(x,y)\ne N_x(x,y)$.

Since the equation is not exact, we must make it exact with the IFM.

Using $\displaystyle h(x)=\exp[\int^x n(s)ds+k]$ and $\displaystyle n(x)=\frac{-1}{N}(N_x-M_y)$

with $\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4$, $\displaystyle N_x(x,y)=x^2y^4$, and $\displaystyle M_y(x,y)=(5x^3+2)y^4$ we have

$\displaystyle h(x)=\exp \int^x -[\frac{5s^3+s^2+2}{\frac{1}{3}s^3+d_1}]ds$.

# Problem R*3.3 - Find an N1-ODE of the form (1) p. 13-2 and the First Integral, Phi

From Mtg 13-4

## Given

Eq. (1) p.13-2 is $\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$ or $\displaystyle N(x,y)y'+M(x,y)=0$.

Eq. (2) p.13-2 is $\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y)$.

Eq. (1) p.13-3 is $\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x)$.

$\displaystyle a(x) = \sin(x^3)$.

$\displaystyle b(x) = \cos(x)$.

$\displaystyle c(x) = \exp(2y)$.

## Find

$\displaystyle 1)$ Find an N1-ODE of the form (1) p. 13-2 that is either exact or can be made exact by the IFM.

$\displaystyle 2)$ Find the first integral $\phi (x,y)=k$.

## Solution

Solved independently.

$\displaystyle 1)$ First solve for $\displaystyle \bar b(x,y)$ and $\displaystyle \bar c(x,y)$.

$\displaystyle \bar b(x,y) = \int^x \cos(s) ds$

$\displaystyle \bar b(x,y) = \sin(x)$

$\displaystyle \bar c(x,y) = \int^x \exp(2y) ds$

$\displaystyle \bar c(x,y) = \frac{1}{2}*\exp(2y)$

Fill Eq. (1) p. 13-2 with the known values and you will obtain

$\displaystyle \sin(x)*\exp(2y)*y'+\frac{1}{2}*\sin(x^3)*\exp(2y)=0$

This equation simplifies to $\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0$.

So, the above equation satisfies the first condition of exactness, $\displaystyle N(x,y)y'+M(x,y)=0$.

However, it fails to satisfy the second condition of exactness, as shown below.

The second exactness condition requires the formula to satisfy the following form: $\displaystyle M_y(x,y)=N_x(x,y)$.

Differentiate the following equations, $\displaystyle M(x,y)= \frac{1}{2}*\sin(x^3)*\exp(2y)$ and $\displaystyle N(x,y)= \sin(x)*\exp(2y)$,

to obtain $\displaystyle M_y(x,y)= \sin(x^3)*\exp(2y)$ and $\displaystyle N_x(x,y)= \cos(x)*\exp(2y)$.

Clearly, $\displaystyle M_y(x,y)\ne N_x(x,y)$.

Since the equation is not exact, we must make it exact with the IFM.

Using $\displaystyle h(x)=\exp[\int^x n(s)ds+k]$ and $\displaystyle n(x)=\frac{-1}{N}(N_x-M_y)$

with $\displaystyle N(x,y)=\sin(x)$, $\displaystyle N_x(x,y)=\cos(x)$, and $\displaystyle M_y(x,y)=0$ we have

$\displaystyle h(x)=\exp[\int^x [\frac{-1}{\sin(s)}(\cos(s)-0)]ds]$.

This simplies to $\displaystyle h(x)= \frac{1}{\sin(s)}$.

Multiply $\displaystyle h(x)= \frac{1}{\sin(s)}$ and $\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0$ which yields

$\displaystyle y'+\frac{1}{2}\frac{\sin(x^3)}{\sin(x)}=0$.

$\displaystyle 2)$ Find $\phi (x,y)=k$.

$\phi_x(x,y)=M(x,y)$

$\phi_y(x,y)=N(x,y)$

Using the above equations, you can integrate with respect to $\displaystyle x$ or $\displaystyle y$ which yields

$\displaystyle \phi(x,y)=\frac{-1}{2x^2}\cos(x^3)+k_3$

$\displaystyle \phi(x,y)=\sin(x)y+k_4$.

Equate the equations above and solve for $\displaystyle k=k_3+k_4$ which is

$\displaystyle \phi(x,y)=\sin(x)y+\frac{1}{2x^2}\cos(x^3)=k$