Question

Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.

Answer 1

The solution of the diferential equation is:

`y(t)=(1)/(2)cos(t)- (1)/(2)e^(t)+(t)/(2) e^(t)`

Explanation:

Given
`y`; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[
`te^(t)`]

ℒ[y"]+ℒ[y]=ℒ[
`te^(t)`]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[
`te^(t)`]=
`(1)/((s-1)^(2))`

So, the transformation is equal to:

s²·Y(s)+Y(s)=
`(1)/((s-1)^(2))`

(s²+1)·Y(s)=
`(1)/((s-1)^(2))`

Y(s)=
`(1)/((s^(2)+1)(s-1)^(2))`

To be able to separate in terms, we use the partial fraction method:

`(1)/((s^(2)+1)(s-1)^(2))=(As+B)/(s^(2)+1) +(C)/(s-1)+(D)/((s-1)^2)`

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)=
`(s)/(2(s^(2) +1)) -(1)/(2(s-1)) +(1)/(2(s-1)^(2))`

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[
`(s)/(2(s^(2) +1))`]-ℒ⁻¹[
`(1)/(2(s-1))`]+ℒ⁻¹[
`(1)/(2(s-1)^(2))`]

`y(t)=(1)/(2)cos(t)- (1)/(2)e^(t)+(t)/(2) e^(t)`