JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Differential Equations

\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}; y (1) = 1

; then a value of

x

satisfying

y (x) = e

is:

Answer: (C)

Solution

This homogeneous equation yields to the substitution

y = v x

, giving

v + x \frac{d v}{d x} = \frac{v}{1 + v ^{2}}

, so

x \frac{d v}{d x} = \frac{- v ^{3}}{1 + v ^{2}}

. Separating,

\frac{1 + v ^{2}}{v ^{3}} d v = - \frac{d x}{x}

, and integrating gives

- \frac{1}{2 v ^{2}} + ln v = - ln x + c

. With

v = \frac{y}{x}

and

y (1) = 1

the constant is

c = - \frac{1}{2}

. Setting

y = e

then leads to

- \frac{x ^{2}}{2 e ^{2}} + 1 = - \frac{1}{2}

, so

x^{2} = 3 e^{2}

and

x = 3 e

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