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

 

 

 

 
 
 
 
 

Answers (1)

Givenxdy -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

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