Use the Laplace transform to solve the given integral equation

$(s)=2{t}^{2}+{\int}_{0}^{t}\mathrm{sin}[2(t-\tau )]x(\tau )d\tau $

Dolly Robinson
2020-11-02
Answered

Use the Laplace transform to solve the given integral equation

$(s)=2{t}^{2}+{\int}_{0}^{t}\mathrm{sin}[2(t-\tau )]x(\tau )d\tau $

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Corben Pittman

Answered 2020-11-03
Author has **83** answers

Step1

Given

$(s)=2{t}^{2}+{\int}_{0}^{t}\mathrm{sin}[2(t-\tau )]x(\tau )d\tau $

Step2

Taking Laplace transform on both sides of the equation

$L\{x(t)\}=L\{2{t}^{2}+{\int}_{0}^{t}\mathrm{sin}[2(t-\tau )]x(\tau )d\tau \}$

By using linearity property of Laplace transform

$L[af(t)+bg(t)]=aL\{f(t)\}+bL\{f(t)\}$

where a, b are constants.$L\{x(t)\}=2L\left\{{t}^{2}\right\}+L\{{\int}_{0}^{t}sin[2(t-\tau )]x(\tau )d\tau \}$

Step3

By using convolution theorem

$L\{{\int}_{0}^{t}\mathrm{sin}[2(t-\tau )]x(\tau )d\tau \}=L\{\mathrm{sin}(2t)\}L\{x(t)\}$

Therefore,

$L\{x(t)\}=2L\left\{{t}^{2}\right\}+L\{\mathrm{sin}(2t)\}L\{x(t)\}$

Use the formula:

$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},L\{\mathrm{sin}at\}=\frac{a}{{s}^{2}+{a}^{2}}$

It gives

$x(s)=2\left(\frac{2!}{{s}^{2+1}}\right)+\left(\frac{2}{{s}^{2}+{2}^{2}}\right)x(s)$

$x(s)=\frac{4}{{s}^{3}}+\frac{2}{{s}^{2}+4}x(s)$

Solve above equation for x(s)

$x(s)-\frac{2}{{s}^{2}+4}x(s)=\frac{4}{{s}^{3}}$

$(1-\frac{2}{{s}^{2}+4})x(s)=\frac{4}{{s}^{3}}$

$\left(\frac{{s}^{2}+2}{{s}^{2}+4}\right)x(s)=\frac{4}{{s}^{3}}$

Therefore,

$x(s)=\frac{4({s}^{2}+4)}{{s}^{3}({s}^{2}+2)}$

Step 4

By using partial fractions:

$\frac{4({s}^{2}+4)}{{s}^{3}({s}^{2}+2)}=\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}$

Taking inverse Laplace transform on both the sides.

${L}^{-1}\left\{\frac{4({s}^{2}+4)}{{s}^{3}({s}^{2}+2)}\right\}={L}^{-1}\{\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}\}$

Using linearity property of Inverse Laplace transform,

${L}^{-1}\{x(s)\}=-2{L}^{-1}\left\{\frac{1}{s}\right\}+8{L}^{-1}\left\{\frac{1}{{s}^{3}}\right\}+2{L}^{-1}\left\{\frac{s}{{s}^{2}+2}\right\}$

Step 5

Use formulae for Inverse Laplace transform,

Given

Step2

Taking Laplace transform on both sides of the equation

By using linearity property of Laplace transform

where a, b are constants.

Step3

By using convolution theorem

Therefore,

Use the formula:

It gives

Solve above equation for x(s)

Therefore,

Step 4

By using partial fractions:

Taking inverse Laplace transform on both the sides.

Using linearity property of Inverse Laplace transform,

Step 5

Use formulae for Inverse Laplace transform,