Question

5b. Solve the differential equations using Laplace transforms

y"+y' - 2y = x², y(0) = 0 and y'(0) = 0

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Answer 1

`y(x)=-(3)/(4)-(1)/(2)x-(1)/(2)x^2+(2)/(3)e^(-2x)+(1)/(12)e^x`

Explanation:

`y''+y'-2y=x^2,\: y(0)=0,\: y'(0)=0\\\\\mathcal{L}\{y''\}+\mathcal{L}\{y'\}-2\mathcal{L}\{y\}=\mathcal{L}\{x^2\}\\\\s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-2Y(s)=(2)/(s^3)\\ \\s^2Y(s)+sY(s)-2Y(s)=(2)/(s^3)\\ \\(s^2+s-2)Y(s)=(2)/(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)+(B)/(s^2)+(C)/(s^3)+(D)/(s+2)+(E)/(s-1)\\\\2=s^2(s+2)(s-1)A+s(s+2)(s-1)B+(s+2)(s-1)C+s^3(s-1)D+s^3(s+2)E\\\\2=s^(4) A + s^(4) D + s^(4) E + s^(3) A + s^(3) B + 2 s^(3) D - s^(3) E - 2 s^(2) A + s^(2) B + s^(2) C - 2 s B + s C - 2 C\\\\2=s^(4) \left(A + D + E\right) + s^(3) \left(A + B + 2 D - E\right) + s^(2) \left(- 2 A + B + C\right) + s \left(- 2 B + C\right) - 2 C`

Solve for each constant

`\begin{cases} A + D + E = 0\\A + B + 2 D - E = 0\\- 2 A + B + C = 0\\- 2 B + C = 0\\- 2 C = 2 \end{cases}`

`-2C=2\\C=-1`

`-2B+C=0\\-2B+(-1)=0\\-2B-1=0\\-2B=1\\B=-(1)/(2)`

`-2A+B+C=0\\-2A+(-(1)/(2))+(-1)=0\\-2A-(1)/(2)-1=0\\-2A-(3)/(2)=0\\-2A=(3)/(2)\\ A=-(3)/(4)`

`A+D+E=0\\-(3)/(4)+D+E=0\\D+E=(3)/(4)\\D=(3)/(4)-E`

`A+B+2D-E=0\\-(3)/(4)+(-(1)/(2))+2((3)/(4)-E)-E=0\\-(3)/(4)-(1)/(2)+(3)/(2)-2E-E=0\\-(3)/(4)+1-3E=0\\(1)/(4)-3E=0\\(1)/(4)=3E\\(1)/(12)=E`

`D=(3)/(4)-E\\D=(3)/(4)-(1)/(12)\\D=(9)/(12)-(1)/(12)\\D=(8)/(12)\\D=(2)/(3)`

Take the inverse transform and find the solution to the IVP

`Y(s)=(-(3)/(4))/(s)+(-(1)/(2))/(s^2)+(-1)/(s^3)+((2)/(3))/(s+2)+((1)/(12))/(s-1)\\ \\y(x)=-(3)/(4)-(1)/(2)x-(1)/(2)x^2+(2)/(3)e^(-2x)+(1)/(12)e^x`