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.
