# How to find a general real solution to a third order non-homogeneous differential equation?

I am completely stuck with this non homogenous differential equation:

$x'''+(\alpha + \beta)x'' + (1+ \alpha\beta)x' = sin(2t)$

I have to show that the general real solution to aforementioned equation holds the form of

$x(t) = c_1+c_2e^{at}\cos(\omega t) + c_3e^{at}\sin(\omega t) + A\cos (2t) + B\sin (2t)$

and $a, \omega,$ A and B must have determined real parameters. I know I have to solve this by using the method of undetermined coefficients, but those $\alpha$ and $\beta$ are giving me very complicated solutions that I can't wrap my head around. Any solutions?

EDIT:

Due to possible confusion, I put up a link here with the snippet of how the problem is formulated in my book.

• first thing I would do is reduce the order by using $y(t)=x'(t)$ Sep 7 '18 at 16:00

Letting $A = a_1$ and $B = b_1$ (just for notation's sake):

You have $$x(t) = c1 + c2e^{at}cos(wt) + c3e^(at)sin(wt) + a_1 cos(2t) + b_1sin(2t)$$ Deriving three times, we can get $$\begin{split} x'(t) &= \left(ac_3-c_2w\right)\mathrm{e}^{at}\sin\left(wt\right)\\ &+\left(c_3w+ac_2\right)\mathrm{e}^{at}\cos\left(wt\right)-2a_1\sin\left(2t\right)+2b_1\cos\left(2t\right) \end{split}$$

$$\begin{split} x''(t) &= \left(-c_3w^2-2ac_2w + a^2c_3\right)\mathrm{e}^{at}\sin\left(wt\right) \\ &+ \left(-c_2w^2+2ac_3w+a^2c_2\right)\mathrm{e}^{at}\cos\left(wt\right)-4b_1\sin\left(2t\right)-4a_1\cos\left(2t\right) \end{split}$$

$$\begin{split} x'''(t) &= \left(c_2w^3-3ac_3w^2-3a^2c_2w+a^3c_3\right)\mathrm{e}^{at}\sin\left(wt\right)\\ &+\left(-c_3w^3-3ac_2w^2+3a^2c_3w+a^3c_2\right)\mathrm{e}^{at}\cos\left(wt\right)+8a_1\sin\left(2t\right)-8b_1\cos\left(2t\right) \end{split}$$

Now, just plug all three derivatives back in the equation to check if they satisfy the equation, i.e check if $$x'''(t)+(\alpha + \beta)x''(t) + (1+ \alpha\beta)x'(t) = sin(2t)$$

• Sorry but I am confused. Why exactly do you differentiate the answer? Normally, you would write $x(t) = A \cos 2t + B \sin 2t$ and then differentiate it 3 times and plug it into the differential equation Sep 7 '18 at 16:07
• Letting $A = a_1$ and $B = b_1$ (just for notation's sake) Sep 7 '18 at 16:08
• also i think you mean by $A$ as $\alpha$ and $B$ as $\beta$ Sep 7 '18 at 16:09
• That's exactly not what I mean. I wrote the equations just as they are in the book, unless it mixes up $A$ as $\alpha$ and $B$ as $\beta$, which I doubt it does Sep 7 '18 at 16:12
• Okay well you'd have to check if there exists constant $A,B$ that are functions of $\alpha,\beta$ Sep 7 '18 at 16:14

I'll post another approach as @Vasya suggests:

Let $$y = x'$$Then, you get $$y''+(\alpha + \beta)y' + (1+ \alpha\beta)y = sin(2t)$$ Solve the homogenous part $$y''+(\alpha + \beta)y' + (1+ \alpha\beta)y = 0$$ That is get the discriminant $$\Delta = b^2 - 4ac =(\alpha + \beta)^2 - 4 (1+ \alpha\beta) = \alpha^2 + 2\alpha\beta + \beta^2 - 4 - 4\alpha\beta = (\alpha - \beta)^2 - 4$$ which is $$\Delta = (\alpha - \beta - 2)(\alpha - \beta + 2)$$ then $$y_h = k_1 e^{\frac{-(\alpha + \beta) - \sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t} + k_2 e^{\frac{-(\alpha + \beta) +\sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t}$$ Now you need a particular solution of the differential equation $y_p$ which is under the form $$y_p(t) = A\cos(2t) + B\sin(2t)$$ Just plug them up there to get $A$ and $B$ in terms of $\alpha,\beta$, then the solution is $$y(t) = y_h(t) + y_p(t)$$ that is $$y_{\Delta > 0}(t) = k_1 e^{\frac{-(\alpha + \beta) - \sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t} + k_2 e^{\frac{-(\alpha + \beta) +\sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t} + A\cos(2t) + B\sin(2t)$$ If $\Delta > 0$, then we have the above form. If $\Delta < 0$ , then the above becomes $$y_{\Delta < 0}(t) = k_1 e^{\frac{-(\alpha + \beta) -i \sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t} + k_2 e^{\frac{-(\alpha + \beta) +i\sqrt{(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}t} + A\cos(2t) + B\sin(2t)$$ which could be written as $$y_{\Delta < 0}(t) = \frac{k_1}{2} e^{at} (\cos wt - i \sin wt) + \frac{k_2}{2} e^{at} (\cos wt + i \sin wt) + A\cos(2t) + B\sin(2t)$$ where $$w = \frac{\sqrt{-(\alpha - \beta - 2)(\alpha - \beta + 2)}}{2}$$ and $$a = \frac{-(\alpha + \beta)}{2}$$ or even more you could factor the above as $$y_{\Delta < 0}(t) = c_1e^{at} \cos wt + c_2e^{at} \sin wt + A\cos(2t) + B\sin(2t)$$ where $c_1 = \frac{k_1+k_2}{2}$ and $c_2 = -\frac{k_1 - k_2}{2}$. If $\Delta = 0$, then we have If $\Delta > 0$, then we have the above form. If $\Delta < 0$ , then the above becomes $$y_{\Delta = 0}(t) = c_1 e^{\frac{-(\alpha + \beta)}{2}t} + k_2 te^{\frac{-(\alpha + \beta) }{2}t} + A\cos(2t) + B\sin(2t)$$ You should work with all three cases, i'll show you how to work with one to get $x(t)$ back (assuming $\Delta < 0$) $$x(t) = \int y_{\Delta < 0}(t) \ dt = c_0 + c_1 \int e^{at} \cos wt + c_2 \int e^{at} \sin wt + \int A\cos(2t) + \int B\sin(2t)$$ Using integration by parts twice you can easily show that $$\int e^{at} \cos wt = \dfrac{\mathrm{e}^{at}\left(w\sin\left(wt\right)+a\cos\left(wt\right)\right)}{w^2+a^2}$$ also $$\int e^{at} \sin wt = \dfrac{\mathrm{e}^{at}\left(a\sin\left(wt\right)-w\cos\left(wt\right)\right)}{w^2+a^2}+C$$ Also $$\int \cos 2t = \frac{1}{2} \sin t$$ and $$\int \sin 2t =- \frac{1}{2} \cos t$$ Plug everything back up and you'd get the form you're looking for.