JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 9)
Let $$y'\left( x \right) + y\left( x \right)g'\left( x \right) = g\left( x \right),g'\left( x \right),y\left( 0 \right) = 0,x \in R,$$ where $$f'(x)$$ denotes $${{df\left( x \right)} \over {dx}}$$ 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
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Explanation
It is given that
$$y'(x) + y(x)g'(x) = g(x)g'(x)$$
$$ \Rightarrow {e^{g(x)}}y'(x) + {e^{g(x)}}g'(x)y(x) = {e^{g(x)}}g(x)g'(x)$$
$$ \Rightarrow {d \over {dx}}(y(x){e^{g(x)}} )= {e^{g(x)}}g(x)g'(x)$$
Therefore, $$y(x) = {e^{g(x)}} = \int {{e^{g(x)}}g(x)g'(x)dx} $$
$$ = \int {{e^t}t\,dt} $$ [where g(x) = t]
$$ = (t - 1){e^t} + c$$
Therefore, $$y(x){e^{g(x)}} = (g(x) - 1){e^{g(x)}} + c$$
Substituting $$x = 0 \Rightarrow 0 = (0 - 1) \times 1 + c \Rightarrow c = 1$$
Substituting $$x = 2 \Rightarrow y(2) \times 1 = (0 - 1) \times (1) + 1$$
Hence, $$y(2) = 0$$.
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