Find the particular solution of the differential equation $\dfrac{dy}{dx}=\dfrac{xy}{x^{2}+y^{2}}$, given that $y=1$ when $x=0$.

$x^{2}=2y^{2}\log y$ (equivalently $\dfrac{x^{2}}{2y^{2}}=\log y$)

CBSE Class 12 · 2015 · Differential EquationsLong Answer

Find the particular solution of the differential equation

\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}

, given that

y = 1

when

x = 0

Answer: $x^{2} = 2 y^{2} lo g y$ (equivalently $\frac{x ^{2}}{2 y ^{2}} = lo g y$ )

Invert and use

x = v y

(convenient since

y = 1

when

x = 0

\frac{d x}{d y} = \frac{x ^{2} + y ^{2}}{x y}

v + y \frac{d v}{d y} = \frac{v ^{2} + 1}{v} \Rightarrow y \frac{d v}{d y} = \frac{1}{v}

\int v d v = \int \frac{d y}{y} \Rightarrow \frac{v ^{2}}{2} = lo g ∣ y ∣ + C

With

v = \frac{x}{y}

\frac{x ^{2}}{2 y ^{2}} = lo g ∣ y ∣ + C

. Apply

y = 1, x = 0

C = 0

. Hence

x^{2} = 2 y^{2} lo g y

CBSE Class 12 Board · 2015 · Subjective