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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 : \(\frac{2}{e}\)


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


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

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.

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