Question

$\frac{x \: dx + y \: dy}{x \: dy - y \: dx} = \sqrt{( \frac{ {a}^{2} - {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} }) }$
Differential Equation.
solve it.​

Answer 1

Given:

A differential equation

$\bf{\dfrac{xdx+ydy}{xdy-ydx}=\sqrt{\dfrac{a^{2}-x^{2}-y^{2}}{x^{2}+y^{2}}}}$

To Find:

Solution of given differential equation

Solution:

Let x= rcosθ and y= rsinθ

Now,

On squaring and adding x and y, we get

$\longrightarrow\mathrm{x^{2}+y^{2}=(rcos\theta)^{2}+(rsin\theta)^{2}}$

$\longrightarrow\mathrm{x^{2}+y^{2}=r^{2}cos^{2}\theta+r^{2}sin^{2}\theta}$

$\longrightarrow\mathrm{x^{2}+y^{2}=r^{2}(cos^{2}\theta+sin^{2}\theta)}$

$\longrightarrow\mathrm{x^{2}+y^{2}=r^{2}(1)}$

$\longrightarrow\mathrm{\purple{x^{2}+y^{2}=r^{2}}}--------(1)$

On differentiating above equation, we get

$\longrightarrow\mathrm{2xdx+2ydy=2rdr}$

$\longrightarrow\mathrm{\pink{xdx+ydy=rdr}}------(2)$

Now, on dividing y by x, we get

$\longrightarrow\mathrm{\dfrac{y}{x}=\dfrac{\cancel{r}sin\theta}{\cancel{r}cos\theta}}$

$\longrightarrow\mathrm{\dfrac{y}{x}=\dfrac{sin\theta}{cos\theta}}$

$\longrightarrow\mathrm{\dfrac{y}{x}=tan\theta}$

On differentiating above equation we get

$\longrightarrow\mathrm{\dfrac{xdy-ydx}{x^{2}}=sec^{2}\theta d\theta}$

$\longrightarrow\mathrm{xdy-ydx=x^{2}sec^{2}\theta d\theta}$

$\longrightarrow\mathrm{xdy-ydx=(rcos\theta)^{2}sec^{2}\theta d\theta}$

$\longrightarrow\mathrm{xdy-ydx=r^{2}cos^{2}\theta sec^{2}\theta d\theta}$

$\longrightarrow\mathrm{xdy-ydx=r^{2}\cancel{cos^{2}\theta}\dfrac{1}{\cancel{cos^{2}\theta}} d\theta}$

$\longrightarrow\mathrm{\green{xdy-ydx=r^{2}d\theta}}-----(3)$

Now, given differential equation is

$\longrightarrow\mathrm{\dfrac{xdx+ydy}{xdy-ydx}=\sqrt{\dfrac{a^{2}-x^{2}-y^{2}}{x^{2}+y^{2}}}}$

$\longrightarrow\mathrm{\dfrac{xdx+ydy}{xdy-dx}=\sqrt{\dfrac{a^{2}-(x^{2}+y^{2})}{x^{2}+y^{2}}}}$

From (1), (2) and (3), we get

$\longrightarrow\mathrm{\dfrac{rdr}{r^{2}d\theta}=\sqrt{\dfrac{a^{2}-(r^{2})}{r^{2}}}}$

$\longrightarrow\mathrm{\dfrac{dr}{rd\theta}=\dfrac{\sqrt{a^{2}-r^{2}}}{\sqrt{r^{2}}}}$

$\longrightarrow\mathrm{\dfrac{dr}{\cancel{r}d\theta}=\dfrac{\sqrt{a^{2}-r^{2}}}{\cancel{r}}}$

$\longrightarrow\mathrm{\dfrac{dr}{d\theta}=\sqrt{a^{2}-r^{2}}}$

$\longrightarrow\mathrm{\dfrac{dr}{\sqrt{a^{2}-r^{2}}}=d\theta}$

On integrating both the sides, we get

$\longrightarrow\mathrm{\displaystyle\int{\dfrac{dr}{\sqrt{a^{2}-r^{2}}}}\,dr=\int\,d\theta}$

$\longrightarrow\mathrm{\sin^{-1}\bigg(\dfrac{r}{a}\bigg)=\theta+c}$

$\longrightarrow\mathrm{\dfrac{r}{a}=sin(\theta+c)}$

$\longrightarrow\mathrm{\dfrac{r}{a}=sin\theta cosc+cos\theta sinc}$

On multiplying r with both sides, we get

$\longrightarrow\mathrm{\dfrac{r^{2}}{a}=r(sin\theta cosc+cos\theta sinc)}$

$\longrightarrow\mathrm{\dfrac{r^{2}}{a}=rsin\theta cosc+rcos\theta sinc}$

$\longrightarrow\mathrm{r^{2}=a(rsin\theta cosc+rcos\theta sinc)}$

From (1), we get

$\longrightarrow\mathrm{x^{2}+y^{2}=a(rsin\theta cosc+rcos\theta sinc)}$

On substituting rsinθ by y and rcosθ by x

$\longrightarrow\mathrm{x^{2}+y^{2}=a(ycosc+xsinc)}$

$\red{\boxed{\mathrm{x^{2}+y^{2}=a(ycosc+xsinc)}}}$

Hence, the solution of given differential equation is

x²+y²= a(ycosc+xsinc) where c is any constant

Answer 2

Answer:

Refer the attachment

Hope it helps