figoveck38

2022-11-21

Solve ${y}^{\prime}(t)=\mathrm{sin}(t)+{\int}_{0}^{t}y(x)\mathrm{cos}(t-x)dx$ by Laplace transform

My try:

I applied Laplace transform on both sides of the equation.

$sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{cos(t)\ast y(t)\}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{cos(t)\}\times L\{y(t)\}$

Now, I'm stuck on applying the inverse Laplace transform on (*) to find $y(t)$

My try:

I applied Laplace transform on both sides of the equation.

$sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{cos(t)\ast y(t)\}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{cos(t)\}\times L\{y(t)\}$

Now, I'm stuck on applying the inverse Laplace transform on (*) to find $y(t)$

Zackary Hatfield

Beginner2022-11-22Added 14 answers

Hint. You are on the right track. But please check your results, since from your identity

$sL\{y(t)\}(s)=\frac{1}{{s}^{2}+1}+L\{cos(t)\}(s)\times L\{y(t)\}(s)$

using

$L\{cos(t)\}(s)=\frac{s}{{s}^{2}+1}$

I rather get

$L\{y(t)\}(s)=\frac{1}{{s}^{3}}$

which is now standard to solve.

$sL\{y(t)\}(s)=\frac{1}{{s}^{2}+1}+L\{cos(t)\}(s)\times L\{y(t)\}(s)$

using

$L\{cos(t)\}(s)=\frac{s}{{s}^{2}+1}$

I rather get

$L\{y(t)\}(s)=\frac{1}{{s}^{3}}$

which is now standard to solve.

evitagimm9h

Beginner2022-11-23Added 5 answers

$sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{cos(t)\ast y(t)\}$

$sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{y(t)\}\ast \frac{s}{{s}^{2}+1}$

$L\{y(t)\}(s-\frac{s}{{s}^{2}+1})=\frac{1}{{s}^{2}+1}$

Here you made a mistake I guess

$L\{y(t)\}(\frac{{s}^{3}-s+s}{{s}^{2}+1})=\frac{1}{{s}^{2}+1}$

$L\{y(t)\}=\frac{1}{{s}^{3}}$

$sL\{y(t)\}=\frac{1}{{s}^{2}+1}+L\{y(t)\}\ast \frac{s}{{s}^{2}+1}$

$L\{y(t)\}(s-\frac{s}{{s}^{2}+1})=\frac{1}{{s}^{2}+1}$

Here you made a mistake I guess

$L\{y(t)\}(\frac{{s}^{3}-s+s}{{s}^{2}+1})=\frac{1}{{s}^{2}+1}$

$L\{y(t)\}=\frac{1}{{s}^{3}}$

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

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