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Expert-verified Found in: Page 77 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Question: In Problems 33–40, solve the equation given in:Problem 8.

The solution of the given equation in problem 8 is ${\left(\frac{y+\theta }{y}\right)}^{2}=\theta C$.

See the step by step solution

## Step 1: Concept and definition

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.

## Step 2: Analyzing the given statement

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

$\left({y}^{3}-\theta {y}^{2}\right)d\theta +2{\theta }^{2}ydy=0\text{\hspace{0.17em}}........\mathbf{\left(}1\mathbf{\right)}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}$

## Step 3: Rewriting the equation (1) in the form  dydx=G(yx)

Rewriting the equation (1) as,

$\begin{array}{l}\frac{dy}{d\theta }=-\frac{\left({y}^{2}-\theta y\right)}{2{\theta }^{2}}\\ \frac{dy}{d\theta }=-\frac{1}{2}\left(\frac{{y}^{2}}{{\theta }^{2}}-\frac{y}{\theta }\right)\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

## Step 4: Substituting t=yθ   in equation (2) to transform the equation in the variables t and θ

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,

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

$\frac{dt}{d\theta }=\frac{-t\left(t+1\right)}{2\theta }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

## Step 5: Find the solution of equation (3)  dtdθ=-t(t+1)2θ.

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 $t\left(t+1\right)$,

$\begin{array}{c}-2=A\left(t+1\right)+Bt\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)\\ A=-2\end{array}$

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),

$-2\mathrm{ln}\left(t\right)+2\mathrm{ln}\left(t+1\right)=\mathrm{ln}\text{ }\theta +\mathrm{ln}\text{ }C$

Where, C is the constant of integration.

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

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

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

Hence, the solution of the given equation ${\mathbf{\left(}}{{\mathbit{y}}}^{{\mathbf{3}}}{\mathbf{-}}{\mathbit{\theta }}{{\mathbit{y}}}^{{\mathbf{2}}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{\theta }}{\mathbf{+}}{\mathbf{2}}{{\mathbit{\theta }}}^{{\mathbf{2}}}{\mathbit{y}}{\mathbit{d}}{\mathbit{y}}{\mathbf{=}}{\mathbf{0}}$ is ${\left(\frac{y+\theta }{y}\right)}^{{\mathbf{2}}}{\mathbf{=}}{\mathbit{\theta }}{\mathbit{C}}$. ### Want to see more solutions like these? 