Solve 2nd Order Diff Eq: y''=sin(3t)+4e^t, y(0)=1, y'(1)=0

[Dissonance in E]

Homework Statement

find a general solution to the eq:
y''=sin(3t)+4e^t

Find a particular solution that satisfies
y(0) = 1
y'(1) = 0

Homework Equations

The Attempt at a Solution

ive figured the general solution to be
y(x)= -sin(3t)/9 + 4e^t + c + d
And thus
y'(x) = -cos(3t)/3 + 4e^t +c + d

I know the solution is the general solution + the specific values for c & d.

so we get 2 equations

-sin(3(0))/9 +4e^0 +c +d = 1
4+c+d=1
c+d=-3

-cos(3(1))/3 +4e^1 +c +d = 0
-cos(3)/3 + 4e +c +d =0
c+d = cos(3)/3 -4e

now what? if I try substituting values for c & d i end up eliminating both the c and d terms
eg: c = -3 -d
(-3 -d ) + d = cos(3)/3 -4e
-3 = cos(3)/3 -4e ?

 

[Mark44]

Instead of the way you chose, I would just integrate twice to get y(t).

You have y''(t) = sin(3t) + 4e^t
Integrate once to get y'(t) = -(1/3)cos(3t) + 4e^t + C
Use the fact that y'(1) = 0 to solve for C.
Now integrate y'(t) to get y(t), remembering to add a different constant of integration, say D. Use the initial condition y(0) = 0 to solve for D.