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JEE Advanced 2011Paper 2

Exam: JEE Advanced
Year: 2011
Paper: Paper 2
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2011 · Paper 2 · Mathematics · Differential Equations

Let y^ (x)+y(x) g^ (x)=g(x) g^ (x) y(0)=0, x ∈ R , where f^ (x)...

Let

y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x) y(0)=0, x \in R

, where

f^{\prime}(x)

denotes

\frac{d f(x)}{d x}

and

g(x)

is a given non-constant differentiable function on

R

with

g(0)=g(2)=0

. Then, the value of

y

(2) is...

Correct answer

0

Step-by-step solution

\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) g^{\prime}(x)\mathrm{IF}=

e^{\intg^{\prime}(x) d x}=e^{g(x)}

\therefore

Solution is y

\left(

e^{g(x)}\right)=

\int

g(x)

\cdot

g^{\prime}(x)

\cdot

e^{g(x)} d x+C Put

g(x)=t, g^{\prime}(x) d x=d t

\begin{aligned}& y\left(e^{g(x)}\right)=\intt \cdote^t d t+C \\ = & t \cdote^t-\int1 \cdote^t d t+C=t \cdote^t-e^t+C \\ & y e^{g(x)}=(g(x)-1) e^{g(x)}+C \quad \ldots(\mathrm{i}) \end{aligned}

Given,

y(0)=0, g(0)=g(2)=0\therefore

Eq. (i) becomes,

\begin{aligned}& \quad y(0) \cdote^{g(0)}=(g(0)-1) \cdote^{g(0)}+C \\ & \Rightarrow \quad 0=(-1) \cdot1+C \Rightarrow C=1 \\ & \therefore \quad y(x) \cdote^{g(x)}=(g(x)-1) e^{g(x)}+1 \\ & \Rightarrow \quad y(2) \cdote^{g(2)}=(g(2)-1) e^{g(2)}+1 \\ & \text{ where, } \quadg(2)=0 \\ & \Rightarrow \quad y(2) \cdot1=(-1) \cdot1+1 \Rightarrow y(2)=0 \end{aligned}

$

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