+(−x)t=−ex2∕2 which is linear differential equation IF=e∫−xdx=e−x2∕2 Now, solution of differential equation is, t⋅e−x2∕2=−∫e−x2∕2⋅ex2∕2dx ⇒(‌
1
y+1
)e−x2∕2=−x+c, where c is constant of integration....(iii) Given, y(2)=0 i.e. when x=2, then y=0. From Eq. (iii), e−2=−2+c⇒c=2+e−2 Now, at x=1 ‌
1
y+1
e−1∕2=−1+e−2+2 ⇒‌‌(y+1)=‌
e−1∕2
1+e−2
Now, putting the value of (y+1) in Eq. (i), we get y′(1)=‌