Solve the differential equation: xdy-ydx= x^2+y^2 dx, given that … - Vidyadip

Question

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

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Answer

$\text{xdy}-\text{ydx}=\sqrt{\text{x}^2+\text{y}^2}\text{ dx}$
$\Rightarrow\text{xdy}=\Big[\text{y}+\sqrt{\text{x}^2+\text{y}^2}\Big]\text{ dx}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}\ \dots(1)$
Let $\text{F}(\text{x, y})=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}$
$\therefore\text{F}(\lambda\text{x},\lambda\text{y})=\frac{\lambda\text{x}\sqrt{(\lambda\text{x})^2+(\lambda\text{y}^2)}}{\lambda\text{x}}=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}=\lambda^0.\text{F}(\text{x},\ \text{y})$
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
$\text{y}=\text{vx}$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{y})=\frac{\text{d}}{\text{dx}}(\text{vx})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
Substitution the values of v and $\frac{\text{dy}}{\text{dx}}$ in equation (1), we get
$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}+\sqrt{\text{x}^2+(\text{vx})^2}}{\text{x}}$
$\Rightarrow\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sqrt{1+\text{v}^2}$
$\Rightarrow\frac{\text{dv}}{\sqrt{1+\text{v}^2}}=\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get:
$\log\Big|\text{v}+\sqrt{1+\text{v}^2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\Bigg|\frac{\text{y}}{\text{x}}+\sqrt{1+\frac{\text{y}^2}{\text{x}^2}}\Bigg|=\log|\text{Cx}|$
$\Rightarrow\log\Bigg|\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}\Bigg|=\log|\text{Cx}|$
$\Rightarrow\text{y}+\sqrt{\text{x}^2+\text{y}^2}=\text{Cx}^2$
This is the required solution of the given differential equation.

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