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

Answer:

First problem: $y(x)=\frac{1}{2}+e^x-\frac{1}{2}e^{2x}$

Step-by-step explanation:

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

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

Perform partial fraction decomposition

$Y(s)=\frac{1-3s+s^2}{s(s-1)(s-2)}\\\\Y(s)=\frac{s^2-3s+1}{s(s-1)(s-2)}\\ \\\frac{s^2-3s+1}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}\\\\s^2-3s+1=A(s-1)(s-2)+B(s)(s-2)+C(s)(s-1)$

Solve for each constant

$s^2-3s+1=A(s-1)(s-2)+B(s)(s-2)+C(s)(s-1)\\\\(2)^2-3(2)+1=A(2-1)(2-2)+B(2)(2-2)+C(2)(2-1)\\\\-1=2C\\\\-\frac{1}{2}=C$

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

$s^2-3s+1=A(s-1)(s-2)+B(s)(s-2)+C(s)(s-1)\\\\(0)^2-3(0)+1=A(0-1)(0-2)+B(0)(0-2)+C(0)(0-1)\\\\1=2A\\\\\frac{1}{2}=A$

Take the inverse transform to solve the IVP

$Y(s)=\frac{\frac{1}{2}}{s}+\frac{1}{s-1}+\frac{-\frac{1}{2}}{s-2}\\ \\y(x)=\frac{1}{2}+e^x-\frac{1}{2}e^{2x}$