Solve the third-order initial value problem below using the method of Laplace transforms y'''+3y''−18y'−40y=−120 y(0)=6 y'(0)=45 y"(0)=-21

Question
Laplace transform
asked 2021-03-02
Solve the third-order initial value problem below using the method of Laplace transforms
\(y'''+3y''−18y'−40y=−120\)
\(y(0)=6\)
\(y'(0)=45\)
\(y"(0)=-21\)

Answers (1)

2021-03-03
\(\text{Step 1}\)
\(\text{Given equation,}\)
\(y'''+3y''-18y'-40y=-120\)
\(y(0)=6\)
\(y'(0)=45\)
\(y"(0)=-21\)
\(\text{take Laplace transform of both sides of the equation (1)}\\\)
Step 2 \(L\left\{y'''+3y''-18y'-40y\right\}=L\left\{-120\right\}\)
\(s^3L\left\{y\right\}-s^2y(0)-sy'(0)-y''(0)+3(s^2L\left\{y\right\}-sy(0)-y'(0))\)
\(-18(sL\left\{y\right\}-y(0))-40L\left\{y\right\}=\frac{-120}{s}\)
\(\text{Plug in the initial conditions: }y(0)=6 , y'(0)=45 , y"(0)=-21\)
\(s^3L\left\{y\right\}-s^2\times6-s\times45-(-21)+3(s^2L\left\{y\right\}-s\times6-45)-18(sL\left\{y\right\}-6)-40L\left\{y\right\}=\frac{-120}{s}\)
\(s^3L\left\{y\right\}-6s^2-45s+21+3(s^2L\left\{y\right\}6s-45)-18(sL\left\{y\right\}-6)-40L\left\{y\right\}=\frac{-120}{s}\)
\(s^3L\left\{y\right\}+3s^2L\left\{y\right\}-18sL\left\{y\right\}-40L\left\{y\right\}-6s^2-63s-6=\frac{-120}{s}\)
\(L\left\{y\right\}(s^3+3s^2-18s-40)=\frac{-120}{s}+6s^2+63s+6\)
\(L\left\{y\right\}(s^3+3s^2-18s-40)=\frac{-120+6s^3+63s^2+6s}{s}\)
\(L\left\{y\right\}=\frac{-120+6s^3+63s^2+6s}{s(s^3+3s^2-18s-40)}\)
\(\text{Step 3}\)
\(\text{Resolve into partial fractions,}\)
\(\frac{6s^3+63s^2+6s-120}{s(s^3+3s^2-18s-40)}=\frac{6s^3+63s^2+6s-120}{s(s+2)(s-4)(s+5)}\)
\(=\frac{A}{s}\frac{B}{s+2}\frac{C}{s-4}\frac{D}{s+5}\)
\(6s^3+63s^2+6s-120=A(s+2)(s-4)(s+5)+Bs(s-4)(s+5)+Cs(s+2)(s+5)+Ds(s+2)(s-4)\)
\(\text{put } s=0\)
\(-120=-40A\)
\(A=3\)
\(\text{put } s=-2\)
\(B=2\)
\(\text{put } s=4\)
\(C=6\)
\(\text{put } s=-5\)
\(D=-5\)
\(\text{Then, }\)
\(\frac{6s^3+63s^2+6s-120}{s(s+2)(s-4)(s+5)}=\frac{3}{s}+\frac{2}{s+2}+\frac{6}{s-4}-\frac{5}{s+5}\)
\(\text{Step 4}\)
\(\text{Take the inverse Laplace transform }\)
\(y=L^{-1}\left\{\frac{6s^3+63s^2+6s-120}{s(s^3+3s^2-18s-40)}\right\}\)
\(=L^{-1}\left\{\frac{3}{s}+\frac{2}{s+2}+\frac{6}{s-4}-\frac{5}{s+5}\right\}\)
\(y=3+2e^{-2t}+6e^{4t}-5e^{-5t} \text{ is the required solution }\)
0

Relevant Questions

asked 2020-11-09
Solve the third-order initial value problem below using the method of Laplace transforms
\(y'''-2y"-21y'-18y=-18\)
\(y(0)=2\)
\(y'(0)=7\)
\(y"(0)=95\)
asked 2021-02-09
Solve the initial value problem below using the method of Laplace transforms
\(2ty''-3ty'+3y=6,y(0)=2, y'(0)=-3\)
asked 2020-11-17
Solve the initial value problem below using the method of Laplace transforms
\(y"-35y=144t-36^{-6t}\)
\(y(0)=0\)
\(y'(0)=47\)
asked 2021-02-21
Solve the initial value problem below using the method of Laplace transforms.
\(y"-16y=32t-8e^{-4t}\)
\(y(0)=0\)
\(y'(0)=15\)
asked 2021-01-31
Solve the initial value problem below using the method of Laplace transforms.
\(y"-4y'+40y=225 e^{5t}\)
\(y(0)=5\)
\(y'(0)=31\)
asked 2021-01-10
Solve the following initial value problems using Laplace Transforms:
\(a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,\)
\(y(0=0) , y'(0)=0\)
\(b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,\)
\(y(0=0) , y'(0)=0\)
\(c) 2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t},\)
\(y(0=0) , y'(0)=-2\)
asked 2020-12-16
solve the initial value problem with using laplace
\(Y'+3Y=6e^{-3t}\cos 6t+24\)
\(Y(0)=0\)
asked 2021-03-09
How to solve for third order differential equation of \(y"'-7y'+6y =2 \sin (t)\) using Method of Laplace Transform when \(y(0)=0, y'(0)=0, y"(0)=0\)?
Step by step
asked 2021-02-14
Use Laplace transforms to solve the following initial value problem
\(y"-y'-6y=0\)
\(y(0)=1\)
\(y'(0)=-1\)
asked 2020-11-16
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
\(y"+y=g(t) , y(0)=-4 , y'(0)=0\)
where \(g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}\)
Y(s)-?
...