Question

5c. Solve the differential equations using Laplace transforms

y"+y' - 2y = e-³x, y(0) = 0 and y'(0) = 0​

Answer 1

`y(x)=(1)/(2)e^(-3x)-(2)/(3)e^(-2x)+(1)/(6)e^(x)`

Explanation:

`y''+y'-2y=e^(-3x),\: y(0)=0,\: y'(0)=0\\\\\mathcal{L}\{y''\}+\mathcal{L}\{y'\}-2\mathcal{L}\{y\}=\mathcal{L}\{e^(-3x)\}\\\\s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-2Y(s)=(1)/(s+3) \\ \\s^2Y(s)+sY(s)-2Y(s)=(1)/(s+3)\\ \\(s^2+s-2)Y(s)=(1)/(s+3)\\ \\Y(s)=(2)/((s+3)(s^2+s-2))\\ \\Y(s)=(2)/((s+3)(s+2)(s-1))`

Perform the partial fraction decomposition

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

Solve for each constant

`2=((-2)+2)((-2)-1)A+((-2)+3)((-2)-1)B+((-2)+2)((-2)+3)C\\\\2=-3B\\\\-(2)/(3)=B`

`2=(1+2)(1-1)A+(1+3)(1-1)B+(1+2)(1+3)C\\ \\2=12C\\\\(2)/(12)=C\\ \\(1)/(6)=C`

`2=((-3)+2)((-3)-1)A+((-3)+3)((-3)-1)B+((-3)+2)((-3)+3)C\\ \\2=4A\\\\(2)/(4)=A\\ \\(1)/(2)=A`

Take the inverse transform and solve for the IVP

`Y(s)=((1)/(2))/(s+3)+(-(2)/(3))/(s+2)+((1)/(6))/(s-1)\\\\y(x)=(1)/(2)e^(-3x)-(2)/(3)e^(-2x)+(1)/(6)e^(x)`