# Solve differential equation with unknown coefficients

Problem

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?

• Factorise $\lambda$ thats the first thing to do then you have a quadratic equation that you can solve. Sep 6 '20 at 16:55
• @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?
– Carl
Sep 6 '20 at 17:01
• Of order two Carl 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 $$A,B$$.

• 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.
– Carl
Sep 6 '20 at 17:12
• You're welcome Carl. Have a good day too. @Carl Sep 6 '20 at 17:13
• 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?
– Carl
Sep 7 '20 at 8:16
• I added some lines @Carl Sep 7 '20 at 11:09
• 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 Sep 7 '20 at 11:23