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

Answers (1)

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

$\frac{xdy}{dx} - y = \sqrt{x^2 + y^2} \Rightarrow \frac{xdy}{dx} = \sqrt{x^2 + y^2} + y$

$\Rightarrow \frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x} \quad -(1)$

Let $y = vx$

Differentiating w.r.t $x$

$\Rightarrow \frac{dy}{dx} = x + x\frac{dv}{dx}$

Put in (1)

$\Rightarrow v + x \frac{dv}{dx} =v + \frac{\sqrt{x^2 + v^2x^2} }{x}$

$\Rightarrow v + x \frac{dv}{dx} =v + \sqrt{1 + v^2}$

$\Rightarrow x \frac{dv}{dx} =v + \sqrt{1 + v^2} -v$

$\Rightarrow x \frac{dv}{dx} =\sqrt{1 + v^2}$

$\Rightarrow \frac{dv}{\sqrt{1 + v^2}} =\frac{dx}{x}$

Integrating both sides

$\Rightarrow \int \frac{dv}{\sqrt{1 + v^2}} =\int \frac{dx}{x}$

$\Rightarrow \log \left (v + {\sqrt{1 + v^2}}\right ) =\log x + \log c$

$\Rightarrow \log \left (v + {\sqrt{1 + v^2}}\right ) =\log cx$

Taking substitution to get

$\\y = vx \\ v = \frac{y}{x}$

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

$\Rightarrow y + \sqrt{x^2 + y^2} = cx^2\quad -(2)$

Now given $y = 0$ when $x =1$

from equation (2)

$\\0+ 1 = c \\ c = 1$

Substituting the value of c in (1) we get,

$y + \sqrt{x^2 + y^2} = x^2$

Posted by

Ravindra Pindel

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