Lennie Carroll

## Answered question

2021-01-19

Use Laplace transforms to solve the following initial value problem
$x"-25x=9t$
$x\left(0\right)={x}^{\prime }\left(0\right)=0$

### Answer & Explanation

Nichole Watt

Skilled2021-01-20Added 100 answers

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

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