Question

Use the Laplace Transform to solve the equation: `y''-3y'+2y=u_5(t)`; `y(0)=0`, `y'(0)=0`.

Answer

`y''-3y'+2y=u_5(t)`, `y(0)=0,y'(0)=0`.

Take Laplace transform of both sides:

`L(y''-3y'+2y)=L(u_5(t))`.

`L(y'')-3L(y')+2L(y)=L(u_5(t))`.

Using the following formulas: `L(y'')=s^2Y(s)-sy(0)-y'(0)`, `L(y')sY(s)-y(0)`,

`L(u_5(t))=e^(-5s)/s`, where `L(y)=Y(s)`, we can rewrite equation as follows:

`s^2Y(s)-sy(0)-y'(0)-3(sY(s)-y(0))+2Y(s)=e^(-5s)/s`.

Applying initial conditions gives:

`s^2Y(s)-s*0-0-3(sY(s)-0)+2Y(s)=e^(-5s)/s`.

Or

`(s^2-3s+2)Y(s)=e^(-5s)/s`.

`Y(s)=e^(-5s)/(s(s^2-3s+2))=e^(-5s)/(s(s-1)(s-2))`.

In order to determine solution we must find inverse Laplace Transform.

Partial fraction decomposition for `1/(s(s-1)(s-2))` is `A/s+B/(s-1)+C/(s-2)`.

So,

`A/s+B/(s-1)+C/(s-2)=(A(s-1)(s-2)+Bs(s-2)+Cs(s-1))/(s(s-1)(s-2))=`

`=(A(s^2-3s+2)+B(s^2-2s)+C(s^2-s))/(s(s-1)(s-2))=((A+B+C)s^2+(-3A-2B-C)s+(2A))/(s(s-1)(s-2))`.

It must equal `1/(s(s-1)(s-2))`, so we have following system:

`{(A+B+C=0),(-3A-2B-C=0),(2A=1):}`.

From last equation `A=1/2`.

Add up first two equations: `-2A-B=0` or `B=-1`.

From first equation `C=-A-B=-1/2+1=1/2`.

So, `1/(s(s-1)(s-2))=1/2*1/s-1/(s-1)+1/2*1/(s-2)`.

Finally,

`y(t)=L^-1(Y(s))=L^-1(e^(-5s)/(s(s-1)(s-2)))=L^-1(e^(-5s)(1/2*1/s-1/(s-1)+1/2*1/(s-2)))=`

`=1/2L^-1(e^(-5s)/s)-L^-1(e^(-5s)/(s-1))+1/2L^-1(e^(-5s)/(s-2))=1/2u_5(t)-u_5(t)e^(t-5)+1/2u_5(t)e^(2(t-5))=`

`=1/2u_5(t)(1-2e^(t-5)+e^(2t-10))`.

Answer: `y(t)=1/2u_5(t)(1-2e^(t-5)+e^(2t-10))`.