Question

5. Solve the differential equations using Laplace transforms

y'' − 3y' + 2y = 1, y(0) = 1, y'(0) = 0

y'' + y' − 2y = x², y(0) = 0 and y'(0) = 0

y'' + y' − 2y = e⁻³ˣ, y(0) = 0 and y'(0) = 0

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Answer 1

First problem:
`y(x)=(1)/(2)+e^x-(1)/(2)e^(2x)`

Explanation:

Solve for Y(s) by taking the transform of every term

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

Perform partial fraction decomposition

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

Solve for each constant

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

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

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

Take the inverse transform to solve the IVP

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