## Question

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

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)).