Question

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

Answer 1

Answer:

The solution of the diferential equation is:

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$

Step-by-step explanation:

Given $y" + y = te^{t}$; 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[$te^{t}$]=$\frac{1}{(s-1)^{2}}$

So, the transformation is equal to:

s²·Y(s)+Y(s)=$\frac{1}{(s-1)^{2}}$

(s²+1)·Y(s)=$\frac{1}{(s-1)^{2}}$

Y(s)=$\frac{1}{(s^{2}+1)(s-1)^{2}}$

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

$\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}$

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)=$\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[$\frac{s}{2(s^{2} +1)}$]-ℒ⁻¹[$\frac{1}{2(s-1)}$]+ℒ⁻¹[$\frac{1}{2(s-1)^{2}}$]

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$