Solve the following Ivp using any method y 4y te2t y0 3 y

Solve the following Ivp, using any method, {y\" - 4y = te^2t y(0) = -3 y\'(0) = 1 Alternatively: apply the LT. to the IVP and solve for L{y}. Solution:

Solution

Let us solve IVP using Laplace Transformations

y\"-4y=te^t

take laplace transformations on both side

L{y\"}-4L{y}=L{te^t}

we have the formula

L(y\'\')=s^2Y(s)-sy(0)-y\'(0)

L(y(t))=Y(s)

L(t^ne^at)=n!/(s-a)^n+1

apply these formula in the problem

we get s^2Y(s)-sy(0)-y\'(0)-4Y(s)=1/(s-2)^2

plug in y(0)=-3 y\'(0)=1

s^2Y(s)+3s-1-4Y(s)=1/(s-2)^2

(s^2-4)Y(s)+3s-1    =1/(s-2)^2

(s^2-4)Y(s)             =1/(s-2)^2+1-3s                      Taking lcd

                            ={1+(1-3s)(s-2)^2} /(s-2)^2

               Y(s)       ={1+(1-3s)(s-2)^2} /(s-2)^2(s^2-4)               we can write s^2-4 =(s+2)(s-2)

               Y(s)       ={1+(1-3s)(s-2)^2} /(s+2)(s-2)^3

using partial fractions

{1+(1-3s)(s-2)^2} /(s+2)(s-2)^3 =A/s+2 +B/(s-2) +c/(s-2)^2+D/(s-2)^3

next we have to find the values of A B C D

Take lcd and cancel out the common denominator

{1+(1-3s)(s-2)^2} = A(s-2)^3+B(s+2)(s-2)^2+C(s+2)(s-2)+D(s+2)

putting s=2 we get D=1/4

putting s=-2 we get A=-113/64

put s=0 and s=1 we get equations, solving those two equations, we get B=-79/64 C=-4/64

so we get Y(s) = (-113/64)/s+2 +(-79/64)/(s-2) +(-4/64)/(s-2)^2+(16/64)/(s-2)^3

taking common denomiator 64 outside

    Y(s) = (1/64) {-113/(s+2) -79/(s-2) -4/(s-2)^2+16/(s-2)^3}

Take inverse laplace transform on both side                  

L^-1{Y(s) = y(t) L^-1{1/s+2)}= e^-2t L^-1{1/(s-2)} =e^2t L^-1{1/(s-2)^2}=te^2t     L^-1{1/(s-2)^3}=(1/2)t^2e^2t

              

y(t) =1/64 {-113e-2t -79e2t-4te2t-8t^2e2t}