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

Solve the initial value problem below using the method of Laplace transforms
$y"-35y=144t-{36}^{-6t}$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=47$
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Solution:
The differential equation is
$y"-35y=144t-{36}^{-6t}$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=47$
Apply Laplace transforms:
$L\left\{y"-35y\right\}=L\left\{144t-{36}^{-6t}\right\}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-36Y\left(s\right)=\frac{144}{{s}^{2}}-\frac{36}{s+6}$
$\left({s}^{2}-36\right)Y\left(s\right)-47=\frac{144}{{s}^{2}}-\frac{36}{s+6}$
$Y\left(s\right)=\frac{144}{{s}^{2}\left({s}^{2}-36\right)}-\frac{36}{\left(s+6\right)\left({s}^{2}-36\right)}+\frac{47}{\left({s}^{2}-36\right)}$
Decompose each term:
$\frac{144}{{s}^{2}\left({s}^{2}-36\right)}=\frac{A}{s}+\frac{B}{{s}^{2}}+\frac{C}{s+6}+\frac{D}{s-6}$
$=\frac{As\left({s}^{2}-36\right)}{{s}^{2}\left({s}^{2}-36\right)}+\frac{B\left({s}^{2}-36\right)}{{s}^{2}\left({s}^{2}-36\right)}+\frac{C{s}^{2}\left(s-6\right)}{{s}^{2}\left({s}^{2}-36\right)}+\frac{D{s}^{2}\left(s+6\right)}{{s}^{2}\left({s}^{2}-36\right)}$
$=\frac{\left(A+C+D\right){s}^{3}+\left(B-6C+6D\right){s}^{2}-36As-36B}{{s}^{2}\left({s}^{2}-36\right)}$

Step 2
$\frac{144}{{s}^{2}\left({s}^{2}-36\right)}=\frac{-4}{{s}^{2}}-\frac{1}{3\left(s+6\right)}+\frac{1}{3\left(s-6\right)}$
$\frac{36}{\left(s+6\right)\left({s}^{2}-36\right)}=\frac{A}{s+6}+\frac{B}{\left(s+6{\right)}^{2}}+\frac{C}{\left(s-6\right)}$
$=\frac{A\left({s}^{2}-36\right)+B\left(s-6\right)+C\left(s+6{\right)}^{2}}{\left(s+6\right)\left({s}^{2}-36\right)}$
$A=\frac{1}{4},B=3,C=-\frac{1}{4}$
$\frac{36}{\left(s+6\right)\left({s}^{2}-36\right)}=\frac{1}{4\left(s+6\right)}+\frac{3}{\left(s+6{\right)}^{2}}-\frac{1}{4\left(s-6\right)}$
$\frac{47}{\left({s}^{2}-36\right)}=\frac{A}{s+6}+\frac{B}{s-6}$

Conclusion:
Hence, we have