Question: In Problems 33–40, solve the equation given in:

Problem 8.

When the equation is of the form,$\frac{dy}{dx}=G\left(\frac{y}{x}\right)$, then the substitution$v=\frac{y}{x}$transforms the given equation into a separable equation in the variables v and x.

One has to solve the equation given in problem 7, i.e.,

$\left(y^3-\theta y^2\right)d\theta +2{\theta }^{2}ydy=0\mathrm{........}\mathbf{\left(}1\mathbf{\right)}$

Rewriting the equation (1) as,

$\begin{array}{l}\frac{dy}{d\theta }=-\frac{\left(y^2-\theta y\right)}{2{\theta }^2}\\ \frac{dy}{d\theta }=-\frac{1}{2}\left(\frac{y^2}{{\theta }^2}-\frac{y}{\theta }\right)\mathrm{......}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Substitute$t=\frac{y}{\theta }$in equation (2),

So,$dt=\frac{\theta dy-yd\theta }{{\theta }^2}$

${\theta }^{2}dt=\theta dy-yd\theta$

Dividing both sides by $d\theta$,

$\begin{array}{c}{\theta }^2\frac{dt}{d\theta }=\theta \frac{dy}{d\theta }-y\\ {\theta }^2\frac{dt}{d\theta }+y=\theta \frac{dy}{d\theta }\\ \frac{dy}{d\theta }=\theta \frac{dt}{d\theta }+\frac{y}{\theta }\\ \frac{dy}{d\theta }=\theta \frac{dt}{d\theta }+t\end{array}$

Therefore, equation (2) becomes,

$\begin{array}{c}\theta \frac{dt}{d\theta }+t=-\frac{1}{2}\left(t^2-t\right)\\ 2\theta \frac{dt}{d\theta }+2t=-t^2+t\\ 2\theta \frac{dt}{d\theta }=-t^2+t-2t\\ 2\theta \frac{dt}{d\theta }=-t^2-t\end{array}$

$\frac{dt}{d\theta }=\frac{-t\left(t+1\right)}{2\theta }\mathrm{......}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Separate the variables in equation (3),

$\frac{-2}{t\left(t+1\right)}dt=\frac{1}{\theta }d\theta$

Integrating both sides,

$\int \frac{-2}{t\left(t+1\right)}dt=\int \frac{1}{\theta }d\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)$

The left hand side of the equation (4) can be solved by using partial fraction,

Therefore,

$\frac{-2}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1}$

Multiply both sides by $t\left(t+1\right)$,

$\begin{array}{c}-2=A\left(t+1\right)+Bt\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)\\ A=-2\end{array}$

Put t = -1 in equation (5),

$\begin{array}{c}-2=-B\\ B=2\end{array}$

Therefore, equation (4) becomes,

$\int \frac{-2}{t}dt+\int \frac{2}{t+1}dt=\int \frac{1}{\theta }d\theta$

Put t = 0 in equation (5),

$-2\ln \left(t\right)+2\ln \left(t+1\right)=\ln \theta +\ln C$

Where, C is the constant of integration.

$\begin{array}{c}\ln {\left(t+1\right)}^2-\ln \left(t^2\right)=\ln +\ln C\\ \ln \left(\frac{{\left(t+1\right)}^2}{t^2}\right)=\ln \left(\theta C\right)\\ \frac{{\left(t+1\right)}^2}{t^2}=\theta C\cdot \cdot \cdot \cdot \cdot \cdot \left(6\right)\end{array}$

Put$t=\frac{y}{\theta }$in equation (6),

${\left(\frac{y+\theta }{y}\right)}^2=\theta C$

Hence, the solution of the given equation $\mathbf{(}{\mathit{y}}^{\mathbf{3}}\mathbf{-}\mathit{\theta }{\mathit{y}}^{\mathbf{2}}\mathbf{)}\mathit{d}\mathit{\theta }\mathbf{+}\mathbf{2}{\mathit{\theta }}^{\mathbf{2}}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$is ${\left(\frac{y+\theta }{y}\right)}^{\mathbf{2}}\mathbf{=}\mathit{\theta }\mathit{C}$.