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I have to find correct trial solution of this equation: $$y''' + 3y'' + 2y' = t + \cos t$$

Attempted work:

$$r^3 + 3r^2 + 2r = 0$$

$$r(r^2 + 3r + 2) = 0$$

$$r(r + 1)(r + 2) = 0$$

$$r= 0,-1,-2$$

$$y_c(t) = C_1e^{-1t} + C_2e^{-2t} + C_3e^0$$

I am having trouble figuring out the correct trial solution for this question. I am not sure how they got these two answers below. Those are the correct solutions to the question. My guess would be, $$y_p(t) = At + B + C\cos t + D\sin t,$$ but that's not correct. Would anyone be able to explain how these two solutions below are the correct solutions to this question? Thank you.

$$y_p(t) = At + Bt^2 + C \cos t + D\sin t$$

$$y_p(t) = At + Bt^2 + Ct^3 + D\cos t + E \sin t$$

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  • $\begingroup$ I understand now the yp1=t(a+bt) equation. What's the logic behind choosing second yp2? The one with Ct^3. $\endgroup$ – Martin Jan 4 '15 at 18:04
  • $\begingroup$ $y_C=-\frac{1}{2}C_1e^{-2t}-e^{-t}C_2+C_3$ $\endgroup$ – Dr. Sonnhard Graubner Jan 4 '15 at 18:14
  • $\begingroup$ The second one doesn't look right at all. That $t^3$ term will result in a $t^2$ term when it is substituted into the original equation, with nothing to cancel it out. $\endgroup$ – Gerry Myerson Jul 24 '15 at 7:09
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Why not reduce the order of the equation?

$$z=y'$$

$$z''+3z'+2z=t+\cos(t)$$

I think the trial function

$$z=at+b$$

will solve

$$z''+3z'+2z=t$$

and the trial function

$$z=k_{1}\cos(t)+k_{2}\sin(t)$$

will solve

$$z''+3z'+2z=\cos(t)$$

However, I would advise finding the solution to the homogeneous equation prior. It could be the case that we think a solution of a nonhomogeneous equation is actually a solution of the homogeneous.

Once you have solved for $z(t)$ in terms of $t$, you will be able to take an anti-derivative to find $y(t)$

$$y(t)=\int z(t)dt$$

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