# How to solve this trigonometrical differential equation?

I have arrived at the following problem: \begin{align} \ddot{u}_{1}(t) + u_1 (t) = -\dfrac{3}{4}a_0^{3} \cos (t + \beta_0) - \dfrac{1}{4} a_0^{3} \cos(3t + 3\beta_0). \end{align}

Of course we know how to solve the homogenous one and has the solution (I have combined analytically the $$\sin$$ and $$\cos$$ function) \begin{align} u_{1,hom} = a_1 \cos (t+\beta_1). \end{align} But how can I solve the non homogenous one? I have tried demanding the solution to be of the form \begin{align} u_p = A \sin (t+\beta_0) + B \cos(3t+ 3\beta_0) \end{align} to no avail.

• You can get two particular solutions to add together if you know how to find a particular solution to $u''+u=A\cos(\omega t+\phi)$.
– anon
Commented May 27, 2022 at 16:42
• The issue is that the non-homogeneous part has $\cos$. You need a linearly independent guess. Try $A t^k \cos(t + \beta_0)$ for the first part and see what $A$ and $k$ should be. Or use the variation of parameters technique: en.wikipedia.org/wiki/Variation_of_parameters Commented May 27, 2022 at 16:42
• Or $At\sin(t+β_0)$ as the next even function to the even right side $\cos(t+β_0)$. Commented May 27, 2022 at 17:16
• On the perturbation theory for the Duffing oscillator (in this special case) see also math.stackexchange.com/questions/3961162/…, math.stackexchange.com/questions/2013417/… Commented May 28, 2022 at 6:31

$$\ddot{u}_{1}(t) + u_1 (t) = -\dfrac{3}{4}a_0^{3} \cos (t + \beta_0) - \dfrac{1}{4} a_0^{3} \cos(3t + 3\beta_0).$$ Or more simply: $$\ddot{u}_{1}(x) + u_1 (x) = 3K \cos (x) +K \cos(3x).$$ Where $$K=- \dfrac{1}{4} a_0^{3}$$ and $$x= t + \beta_0$$

Then for the DE: $$\ddot{u}_{1}(x) + u_1 (x) = 3K \cos (x)$$

The particular slution should be of the form $$u_p=Ax \sin x$$ And for the other DE: $$\ddot{u}_{1}(x) + u_1 (x) = K \cos(3x).$$ Use the particular soluiton: $$u_p=C \cos (3x)$$

We can use inverse operator method. Let f(D) be an operator that when operating on $$u_1(t)$$ gives $$\frac {d^n}{dt^n}u_1 + k_1\frac {d^{n-1}}{dt^{n-1}}u_1 + .... + k_{n-1}\frac {d}{dt}u_1 + k_nu_1$$. Here, f(D) basically looks like a polynomial on D (the characteristic polynomial of the differential equation).

For example, $$(D^2 + D)(u_1) = \frac {d^2}{dt^2}u_1 + \frac {d}{dt}u_1$$

Now, let $$\frac {1}{f(D)}$$ be an operator such that $$\frac {1}{f(D)}T$$ (where T is a function of t), is a function of t, not containing arbitrary constants which when operated upon by f(D) gives T.

So, we have $$f(D) (\frac {1}{f(D)}T) = T$$.

Consider the case when, $$T = Acos(at + b)$$

We can observe that,

$$(D)T = -Aasin(at+b)$$

$$(D^2)T = -Aa^2cos(at+b)$$

$$(D^3)T = Aa^3sin(at+b)$$

and so on..

Thus, $$(D^{2r})T = A(-a^2)^rcos(at+b)$$. And from here,

$$f(D^2) T = f(-a^2) Acos(at +b)$$. Note that $$f(D^2)$$ is an operator but $$f(-a^2)$$ is a polynomial in $$-a^2$$.

By operating with $$\frac 1{f(D^2)}$$ on both sides,

$$\frac 1{f(D^2)} (f(D^2) T) = \frac 1{f(D^2)} (f(-a^2) Acos(at +b))$$

$$=> T = f(-a^2) \frac 1{f(D^2)} Acos(at +b)$$

(or)

$$\frac 1{f(-a^2)}Acos(at+b) = \frac 1{f(D^2)} Acos(at +b)$$. This is true when $$f(-a^2)$$ is not equal to 0. (Refer to the edit for the case when $$f(-a^2)$$ = 0)

So in our case when $$f(D) = D^2 + D$$,

$$\frac 1{D^2 + D}Acos(at+b) = \frac 1{-a^2 + D} Acos(at +b) = \frac {D + a^2}{D^2 - a^4} Acos(at +b) = \frac {D + a^2}{-a^2 - a^4} Acos(at +b) = (D+a^2)\frac {Acos(at+b)}{(-a^2 - a^4)} = \frac {-Aasin(at+b)}{(-a^2 - a^4)} + \frac {a^2Acos(at+b)}{(-a^2 - a^4)}$$

Thus, $$\frac 1{D^2 + D}Acos(at+b) = \frac {-Aasin(at+b)}{(-a^2 - a^4)} + \frac {a^2Acos(at+b)}{(-a^2 - a^4)}$$. Proof of the validity of manipulations done on the operators can be figured out from the definition of the operators.

So, the above expression gives the particular integral of the term/expression T. Thus, the two separate particular solutions for the two terms can then be added to get a particular solution for the entire expression.

Edit: I realized that the differential equation given in the question corresponds to $$(D^2+1)(u1) = T$$ (where T is some function of t). The solutions for this can be obtained simply by using,

$$\frac 1{f(D^2)} Acos(at +b) = \frac 1{f(-a^2)}Acos(at+b)$$ and there is no need to manipulate the operator, since the powers of D in characteristic polynomial are only even. I have not removed the solution for when $$f(D)$$ is $$(D^2 + D)$$ since it serves as an example of a case where not all powers of D are even.

Note: For this question $$f(-a^2)$$ turns out to be $$(-1^2) + 1 = 0$$. So, that particular solution is no longer valid. In these cases we can use $$\frac 1{f(D^2)} Acos(at +b) = \frac 1{f^{'}(-a^2)}Atcos(at+b)$$ where $$f^{'}(-a^2)$$ is not zero. This result can be proved by using the fact that $$cos(at +b)$$ is the real part of $$e^{i(at +b)}$$ and using $$f(D^2) = (D^2+a^2)g(D^2)$$ where $$g(D^2)$$ is some polynomial in $$D^2$$.