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?
Best Answer
$$\lambda^3+\lambda^2(\alpha+\beta)+\lambda(1+\alpha\beta)=0$$ 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 $A,B$.