maeveoneill
Junior Member
- Joined
- Sep 24, 2005
- Messages
- 93
The question is to solve the initial value problem of 2y" + 5y' +3y =0, y(0) =3, y'(0)= -4
I have come up wit the general solution is y=c1e^(-3x/2) + c2xe^(-x)
=> y' = -3/2c1e^(-3x/2) - c2e^(-x)
y(0) = c1 + c2 =3
y'(0) = -3/2c1 - c2 = -4
i am stuck when trying to find c1 in terms of c2.
i get -2/3 or -1/2.. but the back of my book has c1 as 2 and c2 as 1
can anyone tell me how to ge these values for c1 and c2.
thanks
I have come up wit the general solution is y=c1e^(-3x/2) + c2xe^(-x)
=> y' = -3/2c1e^(-3x/2) - c2e^(-x)
y(0) = c1 + c2 =3
y'(0) = -3/2c1 - c2 = -4
i am stuck when trying to find c1 in terms of c2.
i get -2/3 or -1/2.. but the back of my book has c1 as 2 and c2 as 1
can anyone tell me how to ge these values for c1 and c2.
thanks