Solve the differential equation: $\mathrm{xdy}-\mathrm{ydx}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dx}$, given that y=0 when x=1.

  $\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{OR}$

Solve the differential equation: $(1+{\mathrm{x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}-4{\mathrm{x}}^2=0,$ subject to the initial condition y(0)=0.

Given : $\mathrm{xdy}-\mathrm{ydx}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dx}$

$$\begin{gathered} \Rightarrow \mathrm{xdy}=\left[\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\right]\mathrm{dx} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}} ... \left(1\right) \end{gathered}$$

Let $\mathrm{F}(\mathrm{x}, \mathrm{y})=\frac{\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}}.$

$\therefore \mathrm{F}(\mathrm{\lambda x}, \mathrm{\lambda y})=\frac{\mathrm{\lambda x}\sqrt{{\left(\mathrm{\lambda x}\right)}^2+{\left(\mathrm{\lambda y}\right)}^2}}{\mathrm{\lambda x}}=\frac{\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}}={\mathrm{\lambda }}^0·\mathrm{F}(\mathrm{x}, \mathrm{y})$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

$$\begin{gathered} \mathrm{y}=\mathrm{vx} \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{vx}\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} \end{gathered}$$

Substituting the value of v and $\frac{\mathrm{dy}}{\mathrm{dx}}$ in equation (1), we get :

$$\begin{gathered} \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}+\sqrt{{\mathrm{x}}^2+{\left(\mathrm{vx}\right)}^2}}{\mathrm{x}} \\ \Rightarrow \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\sqrt{1+{\mathrm{v}}^2} \\ \Rightarrow \frac{\mathrm{dv}}{\sqrt{1+{\mathrm{v}}^2}}=\frac{\mathrm{dx}}{\mathrm{x}} \end{gathered}$$

Now, integrate on both sides, if follows

${\sin }^{-1}\left(V\right)=\log \left|x\right|+\log C$

$$\begin{gathered} \log \left|\mathrm{v}+\sqrt{1+{\mathrm{v}}^2}\right|=\log \left|\mathrm{x}\right|+\log \mathrm{C} \\ \Rightarrow \log \left|\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}}\right|=\log \left|\mathrm{Cx}\right| \\ \Rightarrow \log \left|\frac{\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}}\right|=\log \left|\mathrm{Cx}\right| \\ \Rightarrow \mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}={\mathrm{Cx}}^2 \end{gathered}$$

Now, given that y=0 when x=1. Substitute this in above equation,

$$\begin{gathered} \therefore 0+\sqrt{1^2+0^2}=\mathrm{C}{\left(1\right)}^{ } \\ \therefore \mathrm{C}=1 \end{gathered}$$

Again, substitute C=1 in the above equation,

$\therefore \mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=1\left({\mathrm{x}}^2\right)\Rightarrow \mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}={\mathrm{x}}^2$

Therefore, the solution of the given differential equation that y=0 when x=1 is, ${\mathrm{x}}^2=\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$

                                                         OR

The given differential equation can be written as:

$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{y}=\frac{4{\mathrm{x}}^2}{1+{\mathrm{x}}^2}...\left(1\right)$

This is a linear differential equation of the form $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$

$$\begin{gathered} \mathrm{Where} \mathrm{P}=\frac{2\mathrm{x}}{1+{\mathrm{x}}^2} \mathrm{and} \mathrm{Q}=\frac{4\mathrm{x}}{1+{\mathrm{x}}^2} \\ Now \mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{\int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}}={\mathrm{e}}^{\log (1+{\mathrm{x}}^2)}=1+{\mathrm{x}}^2 \end{gathered}$$

solution is

$$\begin{gathered} \mathrm{y}\left(1+{\mathrm{x}}^2\right)=\int 4{\mathrm{x}}^2\mathrm{dx}+\mathrm{C} \\ \mathrm{y}\left(1+{\mathrm{x}}^2\right)=\frac{4{\mathrm{x}}^3}{3}+\mathrm{C} ... \left(2\right) \end{gathered}$$

Given y=0, when x=0

Substituting x=0 and y=0 in (1), we get 

$0=0+\mathrm{C}\Rightarrow \mathrm{C}=0$

Substitute C=0 in the equation (2), it follows

$\mathrm{y}(1+{\mathrm{x}}^2)=\frac{\left(4{\mathrm{x}}^3\right)}{3}+0\Rightarrow \mathrm{y}=\frac{\left(4{\mathrm{x}}^3\right)}{3(1+{\mathrm{x}}^2)}$

Which is the required answer.