Differential Equations Part 1

Given, ‌

dy

dx

=(y+1)[(y+1)ex2∕2−x]...(i)
⇒‌‌‌

dy

dx

=(y+1)2ex2∕2−x(y+1)
⇒‌‌‌

dy

dx

+x(y+1)=(y+1)2ex2∕2
⇒‌‌‌

1

(y+1)2

⋅‌

dy

dx

+‌

1

y+1

⋅x=ex2∕2...(ii)
Let ‌

1

y+1

=t⇒‌

−1

(y+1)2

⋅‌

dy

dx

=‌

dt

dx

⇒‌‌−‌

dt

dx

+tx=ex2∕2
⇒‌‌‌

dt

dx

+(−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)=‌

e−1∕2

1+e−2

(‌

e−1∕2

1+e−2

⋅e1∕2−1)
=‌

e−1∕2

1+e−2

(‌

1

1+e−2

−1)
=‌

e−1∕2

1+e−2

(‌

1−1−e−2

1+e−2

)
=‌

−e−5∕2

(1+e−2)2

=‌

−e−5∕2

(e2+1)2 e4

=‌

−e3∕2

(e2+1)2