initial value problem using laplace transform y"+ky'-2k²y=0

Question

Answer is 2e^kt

Comments

Laplace method requires initial conditions to be given. None have been provided.

However, you have provided the solution, which implies that the required initial conditions can be calculated.

y(0)=2 and y'(0)=2k. Is that correct?

Answer 1

If y(0)=2 and y'(0)=2k, then:

ℒ{y"}=s²Y(s)-sy(0)-y'(0)=s²Y(s)-2s-2k; ℒ{y'}=sY(s)-y(0)=sY(s)-2.

For Y(s), simply write Y.

ℒ{y"+ky'-2k²y}=s²Y-2s-2k+skY-2k-2k²Y=0.

Y can be factored out:

Y(s²+sk-2k²)=2s+4k,

Y(s+2k)(s-k)=2s+4k=2(s+2k),

Y(s-k)=2, Y=2/(s-k). Apply inverse Laplace:

2ℒ^-1{1/(s-k)}=2e^kt is the solution.