Question

Solve the differential equation $\mathit{y}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$by changing from variables role="math" localid="1655272385100" $y,x$to $r,x$ where ${\mathit{y}}^{\mathbf{2}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}$; then $\mathit{y}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{n}\mathbf{\text{'}}\mathbf{-}\mathit{x}$.

Short answer

Answer

The solution of the differential equation is $y=\pm \sqrt{-\frac{r^2}{2}\left(r^2-{\left(rr\text{'}\right)}^2\right)}$.

Step-by-step solution

Given information

A differential equation $yy{\text{'}}^2+2xy\text{'}-y=0$

, where $y^2=r^2-x^2$and $yy\text{'}=rr\text{'}-x$

Homogenous function

Homogenous equation: A homogeneous function of x and y of degree n means the function can be written as ${\mathit{x}}^{\mathbf{n}}\mathit{f}\left(\frac{y}{x}\right)$

$\mathit{P}\left(x,y\right)\mathit{d}\mathit{x}\mathbf{+}\mathit{Q}\left(x,y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

, where P and Q are homogeneous function of the same degree.

By the change of variable of homogeneous equation,

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{P}\left(x,y\right)}{\mathbf{Q}\left(x,y\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{f}\left(\frac{y}{x}\right) \end{gathered}$$

Substitute$\mathit{y}\mathbf{=}\mathit{x}\mathit{v}$in the equation.

Solve the differential equation by use of homogeneous function

Put the value of y' in the equation as,

$$\begin{gathered} y{\left(\frac{rr\text{'}-x}{y}\right)}^2+2x\left(\frac{rr\text{'}-x}{y}\right)-y=0 \\ \frac{1}{2}{\left(rr\text{'}-x\right)}^2+2x\left(\frac{rr\text{'}-x}{y}\right)-y=0 \\ \frac{1}{y}\left\{{\left(rr\text{'}-x\right)}^2+2x\left(rr\text{'}-x\right)-y^2\right\}=0 \end{gathered}$$

Put $y^2=r^2-x^2$

$$\begin{gathered} {\left(rr\text{'}-x\right)}^2+2x\left(rr\text{'}-x\right)-\left(rr\text{'}-x\right)-\left(r^2-x^2\right)=0 \\ -2x^2-r^2+{\left(rr\text{'}\right)}^2=0......\left(1\right) \end{gathered}$$

Replace by $y=xr$,

$\frac{y}{r}=x$

Put in equation (1) as,

$$\begin{gathered} -2{\left(\frac{y}{r}\right)}^2-r^2+{\left(rr\text{'}\right)}^2=0 \\ -\frac{2y^2}{r^2}-r^2+{\left(rr\text{'}\right)}^2=0 \\ y=\pm \sqrt{-\frac{r^2}{2}\left(r^2-{\left(rr\text{'}\right)}^2\right)} \end{gathered}$$