Answer to Question #311950 in Differential Equations for kxngToooch

Question

Determine the unique solution of the following differential equations by using Laplace transforms:

a. y"(t) - 6y'(t) + 9y(t) = t²e^3t if y'(0) = 6 and y(0) = 2

b. y"(t) - 2y'(t) + 3y(t) = e^-3t if y'(0) = 0 and y(0) = 0

Answer 1

$a) y''(t)-6y'(t)+9y(t)=t^2e^{3t}\\ y'(0)=6, y(0)=2.\\ L\{y''(t)-6y'(t)+9y(t)\}=L\{t^2e^{3t}\}\\ L\{y''(t)\}-6L\{y'(t)\}+9L\{y(t)\}=L\{t^2e^{3t}\}\\ p^2 Y(p)-py(0)-y'(0)-6(pY(p)-y(0))+\\ +9 Y(p)=\frac{2}{(p-3)^3}\\ Y(p)(p^2-6p+9)-2p-6+12=\frac{2}{(p-3)^3}\\ Y(p)(p-3)^2-2(p-3)=\frac{2}{(p-3)^3}\\ Y(p)(p-3)^2=\frac{2}{(p-3)^3}+2(p-3)\\ Y(p)=\frac{2}{(p-3)^5}+\frac{2}{p-3}\\ L^{-1}\{Y(p)\}=L^{-1}\{\frac{2}{(p-3)^5}\}+L^{-1}\{\frac{2}{p-3}\}\\ y(t)=\frac{2}{4!}t^4e^{3t}+2e^{3t}\\ y(t)=\frac{1}{12}t^4e^{3t}+2e^{3t}\\ b) y''(t)-2y'(t)+3y(t)=e^{-3t}\\ y'(0)=0, y(0)=0.\\ L\{y''(t)-2y'(t)+3y(t)\}=L\{e^{-3t}\}\\ L\{y''(t)\}-2L\{y'(t)\}+3L\{y(t)\}=L\{e^{-3t}\}\\ p^2Y(p)-py(0)-y'(0)-2(pY(p)-y(0))+\\ +3Y(p)=\frac{1}{p+3}\\ Y(p)(p^2-2p+3)=\frac{1}{p+3}\\ Y(p)=\frac{1}{(p+3)(p^2-2p+3)}\\ Y(p)=\frac{1}{18}\frac{1}{p+3}-\frac{1}{18}\frac{p-5}{p^2-2p+3}\\ Y(p)=\frac{1}{18}\frac{1}{p+3}-\frac{1}{18}\frac{p-5}{(p-1)^2+2}\\ Y(p)=\frac{1}{18}\frac{1}{p+3}-\frac{1}{18}\frac{p-1-4}{(p-1)^2+2}\\ Y(p)=\frac{1}{18}\frac{1}{p+3}-\frac{1}{18}\frac{p-1}{(p-1)^2+2}+\frac{2}{9}\frac{1}{(p-1)^2+2}\\ L^{-1}\{Y(p)\}=L^{-1}\{\frac{1}{18}\frac{1}{p+3}\}-\\ -L^{-1}\{\frac{1}{18}\frac{p-1}{(p-1)^2+2}\}+L^{-1}\{\frac{2}{9}\frac{1}{(p-1)^2+2}\}\\ y(t)=\frac{1}{18}e^{-3t}-\frac{1}{18}e^t\cos \sqrt{2}t +\frac{2}{9} \frac{1}{\sqrt{2}}e^t \sin \sqrt{2}t\\ y(t)=\frac{1}{18}e^{-3t}-\frac{1}{18}e^t\cos \sqrt{2}t +\frac{\sqrt{2}}{9} e^t \sin \sqrt{2}t\\$