Question

If the solution of the differential equation \( \frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x} \) satisfies \( y(0)=0 \), then the value of \( y(2) \) is 

(1) -1

(2) 1

(3) 0

(4) e

Answer 1

Correct option is (3) 0

\(\text {I.F. }=e^{\int e^x\left(x^2-2\right) d x}\)

\(=e^{\int e^x\left(x^2-2 x+2 x-2\right)} d x\)

\(=e^{e^x\left(x^2-2 x\right)}\)

\(\text {y. } e^{e^x\left(x^2-2 x\right)}\)

\(=\int \mathrm{e}^{e^x\left(x^2-2 x\right)} e^x\left(x^2-2 x\right)\left(x^2-2\right) e^x d x\)

Let \(e^x\left(x^2-2 x\right)=t\)

So, \(y^{e^{x^2\left(x^2-2 x\right)}}=\int \mathrm{e}^t . t \mathrm{dt}\)

\(\text {At } x=0, t=0\)

\( x=2, t=0 \)

\(=t \cdot e^{\prime}-e^t+c\)

\(x=0 ; 0 \cdot 1\)

\(=0-1+c \Rightarrow c=1\)

\(\text {for } x=2 ; y \cdot 1=0-1+1=0\)

\(y(2) = 0\).