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Bernoulli Equation from an Integral Equation | JEE

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

Question

A continuous function

f: R \to R

satisfies

f(x) = (1+x^2)\left[1 + \displaystyle\int_0^x \dfrac{f^2(t)}{1+t^2}\,dt\right]

. Then the value of

f(-2)

is:

0

\dfrac{17}{15}

-\dfrac{17}{15}

\dfrac{15}{17}

correct

Solution

Step 1: At

x = 0

the integral vanishes, so

f(0) = 1

. Write

\dfrac{f(x)}{1+x^2} = 1 + \displaystyle\int_0^x\frac{f^2(t)}{1+t^2}dt

and differentiate both sides:

\frac{(1+x^2)f'(x) - 2x\,f(x)}{(1+x^2)^2} = \frac{f^2(x)}{1+x^2}.

Step 2: Multiply by

(1+x^2)

and write

y = f(x)

\frac{dy}{dx} - \frac{2x}{1+x^2}\,y = y^2 \quad(\text{Bernoulli}).

Step 3: Divide by

y^2

and put

t = -\dfrac{1}{y}

, so

\dfrac{dt}{dx} = \dfrac{1}{y^2}\dfrac{dy}{dx}

\frac{dt}{dx} + \frac{2x}{1+x^2}\,t = 1, \qquad \text{IF} = e^{\int \frac{2x}{1+x^2}dx} = 1+x^2.

Step 4: Solve:

t(1+x^2) = \int (1+x^2)\,dx = x + \frac{x^3}{3} + c \Rightarrow t = -\frac{1}{f} = \frac{x + \frac{x^3}{3} + c}{1+x^2}.

\therefore f(x) = -\frac{1+x^2}{\frac{x^3}{3} + x + c} = -\frac{3(1+x^2)}{x^3 + 3x + 3c}.

Step 5:

f(0) = 1 \Rightarrow 1 = -\dfrac{3}{3c} \Rightarrow c = -1

, so

f(x) = -\frac{3(1+x^2)}{x^3 + 3x - 3}, \qquad f(-2) = -\frac{3(5)}{-8 - 6 - 3} = \frac{-15}{-17} = \frac{15}{17}.

Correct answer: (4)

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