# $\rm \frac{\mathrm{d} y}{\mathrm{d} x}+2xy=e^{-x^{2}}$ if $y(0)=1$ then $y(1)=?$

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$\rm \frac{\mathrm{d} y}{\mathrm{d} x}+2xy=e^{-x^{2}}$ if $y(0)=1$ then $y(1)=?$
1. $\frac{1}{e}$
2. $\frac{3}{e}$
3. $\frac{2}{e}$
4. 0
5. e

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Correct Answer - Option 3 : $\frac{2}{e}$

Concept:

In the first-order linear differential equation;

$\rm {dy\over dx}+Py=Q$, where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

General solution: y × (IF) = ∫ Q(IF) dx

Calculation:

$\rm \frac{\mathrm{d} y}{\mathrm{d} x}+2xy=e^{-x^{2}}$

So comparing the above equation with the linear differential equation

$\rm \Rightarrow I.F=e^{\int 2xdx}$

$\rm\Rightarrow I.F= e^{x^{2}}$

General solution:

$\rm \Rightarrow ye^{x^{2}}=\int e^{-x^{2}}e^{x^{2}}dx$

$\rm\Rightarrow ye^{x^{2}}=x+c$

At x  = 0 , y = 1

$\rm\Rightarrow c=1$

Therefore, $\rm y=\frac{x+1}{e^{x^{2}}}$

$\rm \therefore y(1)=\frac{2}{e}$

Hence, option 3 is correct.