Solve IVP: Undetermined Coefficient Method

[Fallen6]

Summary:: Initial value problem

Solve the given initial-value problem differential equation by undetermined coefficient method.

d^2 x/dt^2 + w^2 x = F sin wt , x(0) = 0, x'(0) = 0

I get the sol = C1 cos wt + C2 sin wt, but i always get 0 when I plug into the equation, anyone can help me please.

[Thread moved from the technical forums so no Homework Template is shown]

 

[Mark44]

Summary:: Initial value problem

Solve the given initial-value problem differential equation by undetermined coefficient method.

d^2 x/dt^2 + w^2 x = F sin wt , x(0) = 0, x'(0) = 0

I get the sol = C1 cos wt + C2 sin wt, but i always get 0 when I plug into the equation, anyone can help me pls.

Your solution, $x(t) = C_1\cos(\omega t) + C_2\sin(\omega t)$ is a solution to the homogeneous problem, but your DE is a nonhomogeneous equation. You need a particular solution $x_p(t)$ such that $x_p''(t) + \omega^2 x_p(t) = F \sin(\omega t)$.

Do you have any examples of working with nonhomogeneous DEs? If not, take a look at this Insights article - How to Solve Nonhomogeneous Linear ODEs using Annihilators (physicsforums.com)

 

[Fallen6]

I still can't get the answer..., can you show the answer with steps please??

 

[berkeman]

I still can't get the answer..., can you show the answer with steps please??

No, you need to do the bulk of the work on your homework problems. Please show us your work that is not getting you to the answer yet -- that will help us to guide you better. Thanks.

 

[Mark44]

I still can't get the answer..., can you show the answer with steps please??

As already mentioned, this isn't what we do here at PF. Ex. 5 in the link I posted earlier shows an example that is similar to your problem.

 

[Fallen6]

I feel sorry for my ignorant =.=

So, I sub t into my general sol x = At cos wt + Bt cos wt

x' = -Atw sin wt + A cos wt + Btw cos wt + B sin wt

x'' = -Atw^2 cos wt + 2Bw cos wt - Btw^2 sin wt - 2Aw sin wt

#After sub in equation

I get 2Bw cos wt - 2Aw sin wt

Am I wrong from the beginning or I misunderstood something??

 

[Mark44]

I feel sorry for my ignorant =.=

So, I sub t into my general sol x = At cos wt + Bt cos wt

x' = -Atw sin wt + A cos wt + Btw cos wt + B sin wt

x'' = -Atw^2 cos wt + 2Bw cos wt - Btw^2 sin wt - 2Aw sin wt

#After sub in equation

I get 2Bw cos wt - 2Aw sin wt

Am I wrong from the beginning or I misunderstood something??

So far so good, but you lost your equation.
The original DE was $x'' + \omega^2x = F\sin(\omega t)$
After substituting for x'' and x, the equation is $2B\omega \cos(\omega t) - 2A\omega \sin(\omega t) = F\sin(\omega t)$. Solve for A and B, and that will be your particular solution.

Add in the particular solution to get the general solution, which will involve terms of $\cos(\omega t)$ and $\sin(\omega t)$. Then use the initial conditions to determine the coefficients of the $\cos(\omega t)$ and $\sin(\omega t)$ terms.

 

[Fallen6]

I have no idea how to solve A and B, it is using the initial condition to solve for A and B ??

 

[Delta2]

I have no idea how to solve A and B, it is using the initial condition to solve for A and B ??

No I think you should solve this equation $2B\omega\cos\omega t-2A\omega\sin\omega t=F\sin\omega t$. For example one solution is $B=0,A=-\frac{F}{2\omega}$. Substitute these values of B and A at the general form of the solution of post #6 and you ll have got the particular solution.

 

[Fallen6]

I donnot how to get the A and B value like A = ?, B = ?, and If I get the value of A and B, sub into general sol if it is equal to F sin wt then it is the final answer right?

 

[Delta2]

No its not the final answer, it is what we call a particular solution.

To find the final answer I THINK you should add the solution to the homogeneous equation and the particular solution.

I am not so sure though I just know how to solve this equation with Laplace/Fourier Transforms, wait for @Mark44 to reply and offer us his light.

 

[Mark44]

I have no idea how to solve A and B, it is using the initial condition to solve for A and B ??

No. Please reread what I wrote earlier, in post #7.

After substituting for x'' and x, the equation is
$2B\omega \cos(\omega t) - 2A\omega \sin(\omega t) = F\sin(\omega t)$. Solve for A and B, and that will be your particular solution.

The part on the left side, above, is the particular solution, or $x_p(t)$. The equation above has to be true for all values of t, so since the part on the right doesn't contain a cosine term, what can you say about B?

After you find A and B, the general solution will be $x(t) = C_1\cos(\omega t) + C_2\sin(\omega t) + At\cos(\omega t) + Bt\sin(\omega t)$, but with A and B replaced by the values you found. Then use your initial conditions to determine $C_1$ and $C_2$.

 

[Fallen6]

So fat this is what I got.

 

[Mark44]

So fat this is what I got.
View attachment 275392
View attachment 275393

Your work looks OK to me, but you wrote $x_p = A\cos(\omega t) + B\sin(\omega t)$. That's a solution to the homogeneous problem (i.e., $x'' + \omega^2x = 0$, not a particular solution to the nonhomogeneous problem. The work following that seems fine, though.
You've done the hard work -- all you have to do is check that your general solution 1) satisfies the two initial conditions, and 2) satisfies the given differential equation.

 

[vela]

You solved for $C_1$ incorrectly.

 

[Fallen6]

OK thank you all, I will correct my mistakes.