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Find the Curve from a Second-Order Differential Equation | JEE

JEE Maths question with a full step-by-step solution.

Question

If a curve

y = f(x)

satisfies

y''x^2 + x(y')^2 = 2xy'

, with

f(0) = 0

and

f'(1) = 1

, then

f(x)

is:

2(x - \tan^{-1}x)

correct

2\big(x - \ln(1+x)^2\big)

2\big(\ln(1+x^2) - x\big)

DNone of these

Solution

Step 1: Rearrange so a known derivative appears:

y''x^2 - 2xy' = -x(y')^2 \Rightarrow \frac{2xy' - x^2y''}{(y')^2} = x.

The left side is exactly

\dfrac{d}{dx}\left(\dfrac{x^2}{y'}\right)

, so

\frac{d}{dx}\left(\frac{x^2}{y'}\right) = x \Rightarrow \frac{x^2}{y'} = \frac{x^2}{2} + c.

Step 2: Solve for

y'

and use

f'(1)=1

y' = \frac{2x^2}{x^2 + 2c}, \qquad 1 = \frac{2}{1+2c} \Rightarrow c = \frac{1}{2}.

\therefore y' = \frac{2x^2}{x^2+1} = 2\left(1 - \frac{1}{1+x^2}\right).

Step 3: Integrate and use

f(0)=0

y = 2\big(x - \tan^{-1}x\big) + c', \qquad f(0)=0 \Rightarrow c' = 0.

\therefore f(x) = 2\big(x - \tan^{-1}x\big).

Correct answer: (1)

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