Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x), \quad y(0)=0, \quad x \in \mathbb{R}$, where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb{R}$ with $g(0)=g(2)=0$. Then the value of $y(2)$ is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Differential Equations

Let

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

, where

f^{'} (x)

denotes

\frac{df ( 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)

Answer: 0

The differential equation is:

y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)

This can be written as:

\frac{d}{d x} [y (x) e^{g (x)}] = e^{g (x)} g (x) g^{'} (x)

Integrating:

y (x) e^{g (x)} = \int e^{g (x)} g (x) g^{'} (x) d x

Let

t = g (x)

, so

d t = g^{'} (x) d x

\int e^{t} \cdot t d t = (t - 1) e^{t} + C

Thus:

y (x) e^{g (x)} = (g (x) - 1) e^{g (x)} + C

Using

y (0) = 0

and

g (0) = 0

0 = (0 - 1) \cdot 1 + C ⟹ C = 1

So:

y (x) e^{g (x)} = (g (x) - 1) e^{g (x)} + 1

x = 2

, with

g (2) = 0

y (2) \cdot 1 = (0 - 1) \cdot 1 + 1 = 0

Therefore,

y (2) = 0

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer