Use Laplace transforms to solve the following initial value problemx"−25x=9tx(0)=x′(0)=0

Step 1
Given:
Given IVP is $x"-25x=9t,x(0)=x^{\prime }(0)=0$
To find solution of IVP we use Laplace transform
step 2
Solution:
Consider,
$x"-25x=9t$
Apply Laplace transform on both side we get
$L\{x^{″}-25x\}=L\left\{9t\right\}$
$\Rightarrow L\left\{x^{″}\right\}-25L\left\{x\right\}=9L\left\{t\right\}$
$\Rightarrow (s^{2}X(s)-sx(0)-x^{\prime }(0))-25X(s)=\frac{9}{s^2}$
Given initial conditions are $x(0)=x^{\prime }(0)=0$ therefore we get
$(s^{2}X(s)-s(0)-0)-25X(s)=\frac{9}{s^2}$
$\Rightarrow (s^2-25)X(s)=\frac{9}{s^2}$
$\Rightarrow X(s)=\frac{9}{s^2(s^2-25)}$
Now we use inverse Laplace transform to find X(t)
$X(t)=L^{-1}\left\{\frac{9}{s^2(s^2-25)}\right\}$
$\Rightarrow X(t)=L^{-1}\left\{\frac{9}{s^2(s+5)(s-5)}\right\}\dots (1)$
Now we take partial fraction
$\frac{9}{s^2(s+5)(s-5)}=\frac{A}{s^2}+\frac{B}{(s+5)}+\frac{C}{(s-5)}$
$\Rightarrow A(s+5)(s-5)+B{s}^2(s-5)+C{s}^2(s+5)=9$
$\Rightarrow A(s^2-25)+B(s^3-5s^2)+C(s^3+5s^2)=9$
$\Rightarrow (B+C)s^3+(A-5B+5C)s^2-25A=9$
comparing both side we get system of equations,
$\{\begin{array}{l}-25A=9\Rightarrow A=-\frac{9}{25}\\ A-5B+5C=0\Rightarrow -5B+5C=\frac{9}{25}\dots (2)\\ B+C=0\dots (3)\end{array}$
$-5B+5C=\frac{9}{25}$
$+5B+5C=0$
$10C=\frac{9}{25}$
Therefore $C=\frac{9}{250}$ substituting value of C in equation 3 we get
$B+C=0\Rightarrow B=-C\Rightarrow B=-\frac{9}{250}$
Therefore Partial fraction of $\frac{9}{s^2(s+5)(s-5)}=-\frac{9}{(25s^2)}-\frac{9}{250(s+5)}+\frac{9}{250(s-5)}$ Hence 1 becomes,
$X(t)=L^{-1}\left\{\frac{9}{s^2(s+5)(s-5)}\right\}$
$=L^{-1}\{-\frac{9}{25s^2}-\frac{9}{250(s+5)}+\frac{9}{250(s-5)}\}$
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