For each of the following initial value problems verify that the accompanying function is a solution:
Function: y = ex+e2x
Verification:
y = ex + e2x
Differentiating both sides we get,
Therefore, ex + e2x is the solution of the differential equation.
Also at x=0, we get y=e0 + e0 which is equal to 1+1=2.
Also at x=0, we get y1=e0+2e0=3.