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Linear Differential Equation with a Function g(x) | JEE

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

Question

Let

y'(x) + \dfrac{g'(x)}{g(x)}\,y(x) = \dfrac{g'(x)}{1 + g^2(x)}

, where

g(x)

is a given non-constant differentiable function on

R

. If

g(1) = y(1) = 1

and

g(e) = \sqrt{2e - 1}

, then

y(e)

equals:

\dfrac{3}{2g(e)}

correct

\dfrac{1}{2g(e)}

\dfrac{1}{g(e)}

\dfrac{2}{3g(e)}

Solution

Step 1: This is linear in

y

with integrating factor

\text{IF} = e^{\int \frac{g'(x)}{g(x)}\,dx} = e^{\ln g(x)} = g(x).

Step 2: Hence

\dfrac{d}{dx}\big(y\,g(x)\big) = g(x)\cdot\dfrac{g'(x)}{1 + g^2(x)}

, and

y\,g(x) = \frac{1}{2}\int\frac{2g(x)g'(x)}{1 + g^2(x)}\,dx = \frac{1}{2}\ln\big(1 + g^2(x)\big) + C.

Step 3: At

x = 1

(

g = 1

y = 1

1 = \dfrac{1}{2}\ln 2 + C \Rightarrow C = 1 - \dfrac{\ln 2}{2}

. Step 4: At

x = e

g^2(e) = 2e - 1

, so

1 + g^2(e) = 2e

y(e)\,g(e) = \frac{1}{2}\ln(2e) + 1 - \frac{\ln 2}{2} = \frac{1}{2}(\ln 2 + 1) + 1 - \frac{\ln 2}{2} = \frac{3}{2}.

\therefore y(e) = \frac{3}{2g(e)}.

Correct answer: (1)

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