If dydx=xyx^2+y^2, y(1)=1; a value of x satisfying y(x)=e is:

Question

If $\dfrac{dy}{dx}=\dfrac{xy}{x^2+y^2}$, $y(1)=1$; a value of $x$ satisfying $y(x)=e$ is:

Accepted Answer

$${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$

Let y = vx

$$ \therefore $$ $${{dy} \over {dx}}$$ = v + x.$${{dv} \over {dx}}$$

$$ \Rightarrow $$  v + x.$${{dv} \over {dx}}$$ = $${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$$ = $${v \over {1 + {v^2}}}$$

$$ \Rightarrow $$ x.$${{dv} \over {dx}}$$ = $${v \over {1 + {v^2}}}$$ - v = $${{v - v - {v^3}} \over {1 + {v^2}}}$$ = $$ - {{{v^3}} \over {1 + {v^2}}}$$

$$ \Rightarrow $$ $$\int {{{1 + {v^2}} \over {{v^3}}}} dv =  - \int {{{dx} \over x}} $$

$$ \Rightarrow $$ $$ - {1 \over {2{v^2}}} + \log v$$ = $$ - \log x + C$$

$$ \Rightarrow $$ $$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x + C$$ ......(1)

putting x = 1, y = 1 we get

$$ \Rightarrow $$ C = $$ - {1 \over 2}$$

From eq. (1)

$$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x  - {1 \over 2}$$

Put y = e

$$ - {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$$ = $$ - \log x  - {1 \over 2}$$

$$ \Rightarrow $$  x2 = 3e2

$$ \Rightarrow $$  x = $$ \pm $$3$$\sqrt e $$

Solutions · 1