Question

3. For each differential equation, find the Laplace transform of the solution y:

3y' + 5y = x^2 , y(0) = 2.

y'' − y = 5e^−4x + 2x, y(0) = y'(0) = 0.

y'' + 3y' + 2y = 5 sin(πx), y(0) = y'(0) = 0.​

Answer 1

Answer:

See below for answer

Step-by-step explanation:

First problem

$3y'+5y=x^2,\: y(0)=2\\\\3\mathcal{L}\{y'\}+5\mathcal{L}\{y\}=\mathcal{L}\{x^2\}\\\\3[sY(s)-y(0)]+5Y(s)=\frac{2}{s^{3}}\\ \\3[sY(s)-2]+5Y(s)=\frac{2}{s^{3}}\\\\3sY(s)-6+5Y(s)=\frac{2}{s^{3}}\\\\(3s+5)Y(s)-6=\frac{2}{s^3}\\ \\(3s+5)Y(s)=\frac{2}{s^3}+6\\\\Y(s)=\frac{2}{s^3(3s+5)}+\frac{6}{3s+5}\\ \\Y(s)=\frac{2+6s^3}{s^3(3s+5)}$

Perform the partial fraction decomposition

$\frac{2+6s^3}{s^3(3s+5)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{3s+5}\\ \\2+6s^3=s^2(3s+5)A+s(3s+5)B+(3s+5)C+s^3D\\\\2+6s^3=3s^3A+5s^2A+3s^2B+5sB+3sC+5C+s^3D\\\\2+6s^3=3s^3A+s^3D+5s^2A+3s^2B+5sB+3sC+5C\\\\2+6s^3=s^3(3A+D)+s^2(5A+3B)+s(5B+3C)+5C$

Solve for each constant

$\begin{cases} 3 A + D = 6\\5 A + 3 B = 0\\5 B + 3 C = 0\\5 C = 2 \end{cases}$

$5C=2\\C=\frac{2}{5}$

$5B+3C=0\\5B+3(\frac{2}{5})=0\\5B+\frac{6}{5}=0\\5B=-\frac{6}{5}\\B=-\frac{6}{25}$

$5A+3B=0\\5A+3(-\frac{6}{25})=0\\5A-\frac{18}{25}=0\\5A=\frac{18}{25}\\A=\frac{18}{125}$

$3A+D=6\\3(\frac{18}{125})+D=6\\\frac{54}{125}+D=6\\D=\frac{696}{125}$

Take the inverse transform and solve for the IVP

$Y(s)=\frac{\frac{18}{125}}{s}+\frac{- \frac{6}{25}}{s^{2}}+\frac{\frac{2}{5}}{s^{3}}+\frac{\frac{696}{125}}{3 s + 5}\\\\y(x)=\frac{18}{125}-\frac{6}{25}x+\frac{1}{5}x^2+\frac{232}{125}e^{-\frac{5}{3}x}$