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Solve the given integral equation for LaTeX: y(t)y ( t ). LaTeX: y(t)+9\displaystyle{\int_{0}^{t}e^{9(t-v)}y(v)\, dv}=\sin(3t)y ( t ) + 9 ∫ 0 t e 9 ( t − v ) y ( v ) d v = sin ⁡ ( 3 t ) Group of answer choices LaTeX: y(t)=3\cos(3t)+9\sin(3t)-9 y ( t ) = 3 cos ⁡ ( 3 t ) + 9 sin ⁡ ( 3 t ) − 9 LaTeX: y(t)=3\cos(3t)+\sin(3t)-3 y ( t ) = 3 cos ⁡ ( 3 t ) + sin ⁡ ( 3 t ) − 3 LaTeX: y(t)=3\cos(3t)+\sin(3t) y ( t ) = 3 cos ⁡ ( 3 t ) + sin ⁡ ( 3 t ) LaTeX: y(t)=3\cos(3t)+9\sin(3t) y ( t ) = 3 cos ⁡ ( 3 t ) + 9 sin ⁡ ( 3 t ) LaTeX: y(t)=\cos(3t)+3\sin(3t)-3

Respuesta :

LammettHash

Looks like the equation is

[tex]y(t)+9\displaystyle\int_0^te^{9(t-v)}y(v)\,\mathrm dv=\sin(3t)[/tex]

Differentiating both sides yields the linear ODE,

[tex]y'(t)+9e^{9(t-t)}y(t)=3\cos(3t)[/tex]

or

[tex]y'(t)+9y(t)=3\cos(3t)[/tex]

Multiply both sides by the integrating factor [tex]e^{9t}[/tex]:

[tex]e^{9t}y'(t)+9e^{9t}y(t)=3e^{9t}\cos(3t)[/tex]

[tex]\left(e^{9t}y(t)\right)'=3e^{9t}\cos(3t)[/tex]

Integrate both sides, then solve for [tex]y(t)[/tex]:

[tex]e^{9t}y(t)=\dfrac1{10}e^{9t}(\sin(3t)+3\cos(3t))+C[/tex]

[tex]y(t)=\dfrac{\sin(3t)+3\cos(3t)}{10}+Ce^{-9t}[/tex]

The given answer choices all seem to be missing C, so I suspect you left out an initial condition. But we can find one; let [tex]t=0[/tex], then the integral vanishes and we're left with [tex]y(0)=0[/tex]. So

[tex]0=\dfrac{0+3}{10}+C\implies C=-\dfrac3{10}[/tex]

So the particular solution is

[tex]y(t)=\dfrac{\sin(3t)+3\cos(3t)-3e^{-9t}}{10}[/tex]

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