T27 7 Solve the initial value problem y 4ye2t with y 0 5 an

T2.7

7. Solve the initial value problem y +4ye^2t with y (0) = 5 and y (0) = -3.

Solution

y\"+4y=24e^2t

D^2+4D=0

m^2+4m=0 === for finding complementary function .

m(m+4)=0 ;

m=0;m=-4

y=c1e^0x +c2e^-4x ; y=c1+c2e^-4x

y(0)=5

x=0;y=5

5=c1+c2e^(-4*0) ;

5=c1+c2-----(1)

y=c1+c2e^-4x

y\'=c2(-4)e^-4x

y\'(0)=-3

x=0;y\'=-3

-3=c2(-4)e^(-4*0)

-3=(-4)c2

as e^0=1

c2=-3/-4 =3/4

c1+c2=5

c1+(3/4)=5

c1=5-(3/4)=20/4-(3/4)=17/4

y=c1+c2e^-4x ; y = (17/4) + (3/4) e^-4x complementary function ;

y = (24e^2t) / (D^2+4D)