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

See below for answer

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)=(2)/(s^(3))\\ \\3[sY(s)-2]+5Y(s)=(2)/(s^(3))\\\\3sY(s)-6+5Y(s)=(2)/(s^(3))\\\\(3s+5)Y(s)-6=(2)/(s^3)\\ \\(3s+5)Y(s)=(2)/(s^3)+6\\\\Y(s)=(2)/(s^3(3s+5))+(6)/(3s+5)\\ \\Y(s)=(2+6s^3)/(s^3(3s+5))`

Perform the partial fraction decomposition

`(2+6s^3)/(s^3(3s+5))=(A)/(s)+(B)/(s^2)+(C)/(s^3)+(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=(2)/(5)`

`5B+3C=0\\5B+3((2)/(5))=0\\5B+(6)/(5)=0\\5B=-(6)/(5)\\B=-(6)/(25)`

`5A+3B=0\\5A+3(-(6)/(25))=0\\5A-(18)/(25)=0\\5A=(18)/(25)\\A=(18)/(125)`

`3A+D=6\\3((18)/(125))+D=6\\(54)/(125)+D=6\\D=(696)/(125)`

Take the inverse transform and solve for the IVP

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