We have the differential equation: $$x^{(3)}(t)+(\alpha+\beta) x''(t)+(1+\alpha\beta)x'(t)=u(t) $$ $\alpha$ and $\beta$ are positive real constants.

With $u(t)=\sin(2t)$ and the assumption $|\alpha-\beta|<2$ use the guessing method to show that the complete real solution is on the form

$$x(t)=c_1+c_2 e^{at}\cos(\omega t)+c_3e^{at}\sin(\omega t)+A\cos(2t)+B\sin(2t) $$

And determine the real parameters, $a, \omega, A, B$.

My progress

Let's first try to find the solution the homogenous differential equation $$x^{(3)}(t)+(\alpha+\beta) x''(t)+(1+\alpha\beta)x'(t)=0 $$

We write up the characteristic polynomial

$$\lambda^3+\lambda^2(\alpha+\beta)+\lambda(1+\alpha\beta)=0 $$

$$\lambda(\lambda^2+\lambda(\alpha+\beta)+(1+\alpha\beta)) =0$$

But, since I don't know the exact values of $\alpha$ and $\beta$ I can't find the roots to the characteristic polynomial, and therefore I can't determine the complete solution to the homogenous differential equation.

I don't know how to get further with this problem form here, but it seems like I'm missing something important that needs to be understood. Or maybe I should use another method instead?

  • $\begingroup$ Factorise $\lambda$ thats the first thing to do then you have a quadratic equation that you can solve. $\endgroup$ – MtGlasser Sep 6 '20 at 16:55
  • $\begingroup$ @Aryadeva After factoring I see that $0$ is definitely a solution. But there must be two more, since the characteristic polynomial is of order 3, right? $\endgroup$ – Carl Sep 6 '20 at 17:01
  • $\begingroup$ Of order two Carl $\endgroup$ – MtGlasser Sep 6 '20 at 17:07

$$\lambda^3+\lambda^2(\alpha+\beta)+\lambda(1+\alpha\beta)=0$$ Factorize $\lambda$ $$\lambda (\lambda^2+\lambda(\alpha+\beta)+(1+\alpha\beta))=0$$ $$\implies \lambda=0$$ And the solution is $y_1=c_1e^{\lambda t}=c_1$ and $$ \lambda^2+\lambda(\alpha+\beta)+(1+\alpha\beta)=0$$ Try to solve the quadratic equation. Now evaluate the discriminant of the quadratic equation and don't forget that $|\alpha -\beta|<2$ $$\Delta =(\alpha +\beta)^2-4(1+\alpha\beta)$$ $$\Delta =(\alpha -\beta)^2-4$$ $$\Delta =(\alpha -\beta)-2)(\alpha -\beta)+2)$$ So that: $$|\alpha -\beta|<2 \implies \Delta <0$$ Then the solutions of the quadratic equation are complex: $$\lambda_{1,2}=\dfrac {-(\alpha+\beta)\pm i\sqrt {|\Delta|}}{2}$$ Now you can deduce the solution to the homogeneous differential equation: $$x(t)=c_1+c_2e^{\lambda_1 t}+c_3e^{\lambda_2 t}$$ You can use Euler's formula in order to reformulate the solution $y(t)$ with sine and cosine functions. $$ \boxed {x(t)=c_1+e^{-(\alpha +\beta)t/2}(c_2 \cos (\sqrt {|\Delta |}t/2)+c_3 \sin (\sqrt {|\Delta |}t/2))}$$ You can easily deduce that: $$\omega =\dfrac 12\sqrt {|\Delta |}=\dfrac 12\sqrt { |(\alpha -\beta)^2-4|}$$ $$a=-\dfrac {(\alpha +\beta)}2$$

For the inhomogeneous part trye $$x(t)=A \sin (2t)+B \cos (2t)$$ $$x'(t)=2A \cos (2t)-2B \sin (2t)$$ $$x''(t)=-4A \sin (2t)-4B \cos (2t)$$ $$x'''(t)=-8A \cos (2t)+8B\sin (2t)$$ Plug these in the DE and find the cosntants $A,B$.

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    $\begingroup$ I made some edits using your hints and I think I'm on the right track. Thank you very much for the help, I will try to take it from here myself. Have a good day, I wish I could give you more than one upvote. $\endgroup$ – Carl Sep 6 '20 at 17:12
  • $\begingroup$ You're welcome Carl. Have a good day too. @Carl $\endgroup$ – MtGlasser Sep 6 '20 at 17:13
  • $\begingroup$ I am trying to find a solution to the inhomogenous solution now and have gotten a very nasty looking equation. I have to determine the coefficients $A$ and $B$ such that the equation is true, but I don't know how to come further. Can you give me some help again? $\endgroup$ – Carl Sep 7 '20 at 8:16
  • $\begingroup$ I added some lines @Carl $\endgroup$ – MtGlasser Sep 7 '20 at 11:09
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    $\begingroup$ Collect all the terms of cosine functions and set it equal to zero. Collect all the terms with sine functions and set it equal to $1$ Then you have a system of two equations in function of $\alpha, \beta$ and solve. @Carl $\endgroup$ – MtGlasser Sep 7 '20 at 11:23

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