Solve the initial value problem below using the method of Laplace transforms y"-35y=144t-36^{-6t} y(0)=0 y'(0)=47

Question
Laplace transform
asked 2020-11-17
Solve the initial value problem below using the method of Laplace transforms
\(y"-35y=144t-36^{-6t}\)
\(y(0)=0\)
\(y'(0)=47\)

Answers (1)

2020-11-18
Solution:
The differential equation is
\(y"-35y=144t-36^{-6t}\)
\(y(0)=0\)
\(y'(0)=47\)
Apply Laplace transforms:
\(L\left\{y"-35y\right\}=L\left\{144t-36^{-6t}\right\}\)
\(s^2Y(s)-sy(0)-y'(0)-36Y(s)=\frac{144}{s^2}-\frac{36}{s+6}\)
\((s^2-36)Y(s)-47=\frac{144}{s^2}-\frac{36}{s+6}\)
\(Y(s)=\frac{144}{s^2(s^2-36)}-\frac{36}{(s+6)(s^2-36)}+\frac{47}{(s^2-36)}\)
Decompose each term:
\(\frac{144}{s^2(s^2-36)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+6}+\frac{D}{s-6}\)
\(=\frac{As(s^2-36)}{s^2(s^2-36)}+\frac{B(s^2-36)}{s^2(s^2-36)}+\frac{Cs^2(s-6)}{s^2(s^2-36)}+\frac{Ds^2(s+6)}{s^2(s^2-36)}\)
\(=\frac{(A+C+D)s^3+(B-6C+6D)s^2-36As-36B}{s^2(s^2-36)}\)
\(A=0 , B=-4 , C=-\frac{1}{3} ,\text{and } D=\frac{1}{3}\)
Step 2
\(\frac{144}{s^2(s^2-36)}=\frac{-4}{s^2}-\frac{1}{3(s+6)}+\frac{1}{3(s-6)}\)
\(\frac{36}{(s+6)(s^2-36)}=\frac{A}{s+6}+\frac{B}{(s+6)^2}+\frac{C}{(s-6)}\)
\(=\frac{A(s^2-36)+B(s-6)+C(s+6)^2}{(s+6)(s^2-36)}\)
\(A=\frac{1}{4} , B=3 , C=-\frac{1}{4}\)
\(\frac{36}{(s+6)(s^2-36)}=\frac{1}{4(s+6)}+\frac{3}{(s+6)^2}-\frac{1}{4(s-6)}\)
\(\frac{47}{(s^2-36)}=\frac{A}{s+6}+\frac{B}{s-6}\)
\(A=-\frac{47}{12} ,\ B=\frac{47}{12}\)
\(\text{Hence, } \frac{47}{(s^2-36)}=-\frac{47}{12(s+6)}+\frac{47}{12(s-6)}\)
Conclusion:
Hence, we have
\(Y(s)=\frac{-4}{s^2}-\frac{1}{3(s+6)}+\frac{1}{3(s-6)}-\frac{1}{4(s+6)}+\frac{3}{(s+6)^2}-\frac{1}{4(s-6)}-\frac{47}{12(s+6)}+\frac{47}{12(s-6)}\)
\(Y(s)=\frac{-4}{s^2}-\frac{9}{2(s+6)}+\frac{4}{(s-6)}+\frac{3}{(s+6)^2}\)
\(\text{Apply inverse Laplace transform: }\)
\(L^{-1}\left\{Y(s)\right\}=-4L^{-1}\left\{\frac{1}{s^2}\right\}-\frac{9}{2}L^{-1}\left\{\frac{1}{(s+6)}\right\}+4L^{-1}\left\{\frac{1}{(s-6)}\right\}+3L^{-1}\left\{\frac{1}{(s+6)^2}\right\}\)
\(y(t)=-4t-\frac{9}{2}e^{-6t}+4e^{6t}+3te^{-6t}\)
\(\text{Therefore, the solution is } y(t)=-4t-\frac{9}{2}e^{-6t}+4e^{6t}+3te^{-6t}\)
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