Question

Solve
d²y/dx²+ y = 1

given y (0) = 1 and y(π/2) = 0 using Laplace transform​

Answer 1

L{dy} + L{2y} = L{12e3t} dt

Now

L{dy} = −y(0)+sY(s)=−3+sY(s) dt

L{2y} = 2Y (s) and L{12e3t} = 12 s−3

Substituting these expressions into the transformed version of the differential equation gives:

[−3+sY(s)]+2Y(s) = 12 s−3

Solving for Y (s) we have (s+2)Y(s)= 12 +3= 3+3s

Therefore

®

s−3 s−3 Y (s) = 3(s + 1)

(s+2)(s−3)

Now, using partial fractions, this last expression can be written in a more convenient form:

Y(s)= 3/5 + 12/5 (s+2) (s−3)

and then, inverting:

y(t)=L−1{Y(s)}=3L−1{ 1 }+12L−1{ 1 } 5s+25s−3

thus

y(t) = 3 e−2tu(t) + 12 e3tu(t)

This is the solution to the given initial value problem.