WBJEE 2012 Math Solved Paper

Let y=emx be the solution of given differential equation,
⇒‌

dy

dx

=memx⇒‌

d2y

dx2

=m2emx
∴‌25‌

d2y

dx2

−10‌

dy

dx

+y=0
⇒‌25m2emx−10memx+emx=0
⇒‌emx(25m2−10m+1)=0
⇒ Auxiliary equation
⇒25m2−10m+1=0
eπx≠0
⇒‌(5m)2−2(5m)×1+1=0
⇒‌(5m−1)2=0
⇒‌m=‌

1

5

,‌

1

5

Since, roots are real and equal
∴ General solution is y=(c1+c2x)ex/5 ...(i)
y(0)=1⇒c1=1
y(1)=2e1/5⇒2e1/5=(c1+c2)e1/5
c1+c2=2⇒c1=1
Putting the value of c1 and c2 in Eq. (i), we get particular solution
y=(1+x)ex/5