A differential equation with undetermined coefficients

[Kaguro]

Homework Statement

Solve the following DE with the method of undetermined coefficients:
y'' + 4y = 2cos(3x)cos(x)

Homework Equations

2cos(3x)cos(x) = cos(4x) + cos(2x)

The Attempt at a Solution

Let's split the particular integral into two parts: yp1 and yp2.
So yp1 is solution for RHS=cos(4x) and yp2 is solution for RHS=cos(2x)

Let yp1 = Acos4x + Bsin4x
(yp1)'' = -16Acos(4x) -16Bsin(4x)

So, -16Acos(4x) - 16Bsin(4x) + Acos(4x) + Bsin(4x) = cos(4x)

So by equating coefficients, we have:
yp1 = (-1/12)cos(4x)

In case of yp2, I have taken:
Let yp2 = Acos(2x) + Bsin(2x)
yp2'' = -4Acos(2x) -4Bsin(2x)

So 0 = cos(2x)
That's not great...

 

[Master1022]

You should look at the LHS and think about what the roots are. Your characteristic equation will be;
$$m^2 + 4 = 0$$

Once you have found your roots (and hence complementary function), you should check whether yp2 is an appropriate choice.

Hope that helps. If not, I am happy to provide more assistance.

 

[Kaguro]

You should look at the LHS and think about what the roots are. Your characteristic equation will be;
$$m^2 + 4 = 0$$

Once you have found your roots (and hence complementary function), you should check whether yp2 is an appropriate choice.

Hope that helps. If not, I am happy to provide more assistance.

That's it!
Thanks very much! You helped me out.

 

[Kaguro]

Do you think this you can help me with another one?

y"+a^2y=cot (ax)

This one too with undetermined coefficients.

If the RHS had been sin or cos then I would have used the complex method... but it is cot.

Cotx doesn't come back if differentiated twice. ..So... any clues?

 

[Master1022]

Do you think this you can help me with another one?
So... any clues?

Sorry, I have just seen your post. Presumably, the assignment has been handed in by now and I apologise for not being able to get back to you in time. Either way, I will still have a think about it.

 

[Kaguro]

Presumably, the assignment has been handed in by now

Yeah, you are right... but I still want to know how to solve this.