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2021-09-23

Using Laplave Transform, evaluate the integro-differential equation
${y}^{″}\left(x\right)+9y\left(x\right)=40{e}^{x};y\left(0\right)=5,{y}^{\prime }\left(0\right)=-2$

izboknil3

Step 1
Solve the given differential equation by Laplace transform.
The given equation is ${y}^{″}\left(x\right)+9y\left(x\right)=40{e}^{x};y\left(0\right)=5,{y}^{\prime }\left(0\right)=-2$
The given equation is
Step 2
Taking the Laplace transform on both sides of given equation.
$L\left\{{y}^{″}\left(x\right)+9y\left(x\right)\right\}=L\left\{40{e}^{x}\right\}$
$L\left\{{y}^{″}\left(x\right)\right\}+9L\left\{y\left(x\right)\right\}=40L\left({e}^{x}\right)$

${s}^{2}\stackrel{―}{y}-5s+2+9\stackrel{―}{y}=\frac{40}{s-1}$
$\stackrel{―}{y}\left({s}^{2}+9\right)=\frac{40}{s-1}+5s-2$
$\stackrel{―}{y}\left({s}^{2}+9\right)=\frac{40+\left(5s-2\right)\left(s-1\right)}{s-1}$
$\stackrel{―}{y}=\frac{5{s}^{2}-7s+42}{\left(s-1\right)\left({s}^{2}+9\right)}\stackrel{˙}{s}\left(2\right)$
Step 3
Solve the R.H.S. of equation (2) by partial fraction.
$\frac{5{s}^{2}-7s+42}{\left(s-1\right)\left({s}^{2}+9\right)}=\frac{A}{s-1}+\frac{Bs+C}{{s}^{2}+9}$
$=\frac{A\left({s}^{2}+9\right)+\left(Bs+C\right)\left(s-1\right)}{\left(s-1\right)\left({s}^{2}+9\right)}$
$=\frac{A{s}^{2}+9A+B{s}^{2}-Bs+Cs-C}{\left(s-1\right)\left({s}^{2}+9\right)}$
$=\frac{\left(A+B\right){s}^{2}+\left(-B+C\right)s+\left(9A-C\right)}{\left(s-1\right)\left({s}^{2}+9\right)}$
Equating the coefficients of same variables.

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