1. ## Second order DE

1)Find the general solution of $x''+x=4 \cos (2t)$. 2)Determine the solution that satisfies $x(0)=0$, $x'(0)=0$.
Attempt: 1)I've found the general solution to the homogeneous DE to be $x_1(t)= \cos (t)$ and $x_2(t)= \sin (t)$.
So I propose a solution to the non-homogeneous DE to be of the form $u_1(t)x_1(t)+u_2(t)x_2(t)$.
$u_1'(t)=\frac{-4 \cos (2t) \sin (t)}{\cos ^2 (t)+ \sin ^2 (t)}=-4 \cos (2t) \sin (t)$. Similarly, $x_2'(t)=-4 \cos (2t) \cos (t)$.
Therefore the general solution to the DE is of the form $x(t)=-4\cos (t) \int \cos (2t) \sin (t) dt -4 \sin (t) \int \cos (2t)\cos (t) dt$. Should I calculate the indefinite integrals or just let it this way? What method would you use to solve the integrals? (I don't see any good u/substitution nor integration by part).
2)I think I should replace t with 0 in my last expression and then equal to 0. But I'm totally unsure about the limits of the integrals. To get $x'(t)$, I would derivate $x(t)$ with respect to t, set $t=0$ and equal $x'(0)=0$ in order to get the constants of integration if they are indefinite integrals? I'm confused on this.

2. Originally Posted by arbolis
1)Find the general solution of $x''+x=4 \cos (2t)$. 2)Determine the solution that satisfies $x(0)=0$, $x'(0)=0$.
Attempt: 1)I've found the general solution to the homogeneous DE to be $x_1(t)= \cos (t)$ and $x_2(t)= \sin (t)$.
So I propose a solution to the non-homogeneous DE to be of the form $u_1(t)x_1(t)+u_2(t)x_2(t)$.
$u_1'(t)=\frac{-4 \cos (2t) \sin (t)}{\cos ^2 (t)+ \sin ^2 (t)}=-4 \cos (2t) \sin (t)$. Similarly, $x_2'(t)=-4 \cos (2t) \cos (t)$.
Therefore the general solution to the DE is of the form $x(t)=-4\cos (t) \int \cos (2t) \sin (t) dt -4 \sin (t) \int \cos (2t)\cos (t) dt$. Should I calculate the indefinite integrals or just let it this way? What method would you use to solve the integrals? (I don't see any good u/substitution nor integration by part).
2)I think I should replace t with 0 in my last expression and then equal to 0. But I'm totally unsure about the limits of the integrals. To get $x'(t)$, I would derivate $x(t)$ with respect to t, set $t=0$ and equal $x'(0)=0$ in order to get the constants of integration if they are indefinite integrals? I'm confused on this.
As you have done I would first solve the homogeneous equation
$x''+x=0$ This gives the complimentary solution
$x(t)=c_1\cos(t)+c_2\sin(t)$
Now since you only have even number derivative the complimentary solution must be of the form
$x_c=A\cos(2t)$
Plugging this into the equation gives

$-4A\cos(2t)+A\cos(2t)=4\cos(2t) \iff -3A=4 \iff A=-\frac{4}{3}$

This gives a solution of

$x=x_c+x_p=c_1\cos(t)+c_2\sin(t)-\frac{4}{3}\cos(2t)$
Now use the I.C to find $c_1$ and $C_2$

3. Thanks for your time and help.
Originally Posted by TheEmptySet
As you have done I would first solve the homogeneous equation
$x''+x=0$ This gives the complimentary solution
$x(t)=c_1\cos(t)+c_2\sin(t)$
Until now I follow you at 100%. I took $c_1=c_2=1$, I thought I could do that but it seems I lose the generality of the solution.
Now since you only have even number derivative
Hmm, what do you mean by that? Since the ODE if of order 2?
the complimentary solution must be of the form
$x_c=A\cos(2t)$
I never saw that before. Is it absolutely required?
Plugging this into the equation gives

$-4A\cos(2t)+A\cos(2t)=4\cos(2t) \iff -3A=4 \iff A=-\frac{4}{3}$

This gives a solution of

$x=x_c+x_p=c_1\cos(t)+c_2\sin(t)-\frac{4}{3}\cos(2t)$
Now use the I.C to find $c_1$ and $C_2$
Ok, this would give me the general solution of the non-homogeneous DE, right? I get that $c_1=\frac{4}{3}$ and $c_2=0$. Is that possible to get a coefficient equal to 0? I guess so because the dimension of the vector space of solution is still 2 in this case, but I just want to be sure.

4. We are looking for a particular solution that is of the form
$x''+x=4\cos(2t)$

This suggests the form of the solution. e.g we want a function x such that it plus its second derivative is $4\cos(2t)$. As we found with the particular solution the only function the only possible solutions are sines and cosines. But an even number of derivatives of a sine function is a sine function. Since we want a cosine function we can assume that the particular solution is made up of only cosines. I hope this helps.

5. Oh ok thank you for the clarification, now I understand why you assumed the complementary (should it be called particular?) solution to be of the form $A \cos (2t)$ which makes sense. Ok I will work out the coefficients $c_1$ and $c_2$ of the non-homogeneous solution.
Though was my method (it is supposed to be variation of parameters) correct? I should also reach the same result via my method, even if it's longer and much harder since I have to evaluate integrals.
Edit: Ok I had calculated $c_1$ and $c_2$ in my above post. So that the general solution is $x(t)=\frac{4}{3} \left [ \cos (t) - \cos (2t) \right ]$, $x'(t)=\frac{4}{3} \left [ 2 \sin (2t) - \sin (t) \right ]$. Is this right? (I should just plug and chug to verify, I know...)
Edit 2: It indeed works... thanks a lot. I wonder if my method works also. And what about those integrals...