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

Engineering

Mathematics

Homogeneous Differential Equation

Question

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

$\sqrt{3} e$

$\frac{1}{2} \sqrt{3} e$

$\sqrt{2} e$

$\frac{e}{\sqrt{2}}$

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Put y = vx

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

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

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

$\Rightarrow \int (\frac{1}{v^{3}} + \frac{1}{v}) dv = \int - \frac{1}{x} dx$

$\Rightarrow \frac{- 1}{2} \frac{1}{v^{2}} + ℓnv = - ℓnx + c$

$\Rightarrow - \frac{x^{2}}{2 y^{2}} = - ℓny + c$

When x = 1, y = 1 then

$- \frac{1}{2} = c$

⇒ x² = y²(1 + 2ℓny)

⇒ x² = e²(3)

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

$So x = \pm \sqrt{3} e$