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balandron [24]
3 years ago
12

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
Mathematics
1 answer:
choli [55]3 years ago
6 0

Looks like the equation is

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

Differentiating both sides yields the linear ODE,

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

or

y'(t)+9y(t)=3\cos(3t)

Multiply both sides by the integrating factor e^{9t}:

e^{9t}y'(t)+9e^{9t}y(t)=3e^{9t}\cos(3t)

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

Integrate both sides, then solve for y(t):

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

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

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

0=\dfrac{0+3}{10}+C\implies C=-\dfrac3{10}

So the particular solution is

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

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Choice A is fairly vague on what the lower and upper boundaries are. What is the smallest number allowed? What about the largest? This isn't clear so it's possible that we could end up with more positive numbers than negative (eg: if we had an interval -10 < x < 110). So choice A is false. A similar issue shows up with choice D.

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