Find the particular solution of the differential equation $$2y\,e^{x/y}\,dx+\left(y-2x\,e^{x/y}\right)dy=0$$ given that $x=0$ when $y=1$.

CBSE Class 12 · 2016 · Differential EquationsLong Answer

Find the particular solution of the differential equation

2 y e^{x / y} d x + (y - 2 x e^{x / y}) d y = 0

given that

x = 0

when

y = 1

Answer: $y = e^{2 - 2 e^{x / y}}$ (equivalently $2 e^{x / y} = 2 - lo g y$ )

Treat

x

as a function of

y

\frac{d x}{d y} = \frac{2 x e ^{x / y} - y}{2 y e ^{x / y}}

, homogeneous. Put

x = v y

\frac{d x}{d y} = v + y \frac{d v}{d y}

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

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

With

v = \frac{x}{y}

2 e^{x / y} = lo g \frac{C}{y}

. Apply

x = 0, y = 1

2 = lo g C \Rightarrow C = e^{2}

. So

2 e^{x / y} = 2 - lo g y

y = e^{2 - 2 e^{x / y}}

CBSE Class 12 Board · 2016 · Subjective