# A differential equation with undetermined coefficients

• Kaguro
In summary: 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.Presumably, the assignment has been handed in by now
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...

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.

Last edited:
Kaguro
Master1022 said:
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.

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?

Kaguro said:
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.

Last edited:
Master1022 said:
Presumably, the assignment has been handed in by now

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

## 1. What is a differential equation with undetermined coefficients?

A differential equation with undetermined coefficients is a type of differential equation where the coefficients of the equation are unknown and need to be determined through a series of steps.

## 2. Why is it important to solve differential equations with undetermined coefficients?

Differential equations with undetermined coefficients are commonly used to model real-world phenomena in fields such as physics, engineering, and economics. Solving them allows us to understand and predict the behavior of these systems.

## 3. What are the steps to solve a differential equation with undetermined coefficients?

The steps to solve a differential equation with undetermined coefficients are:

1. Guess the form of the particular solution based on the type of non-homogeneous term in the equation.
2. Substitute the guessed solution into the original equation and solve for the unknown coefficients.
3. Add the homogeneous solution to the particular solution to get the general solution.

## 4. What is the difference between homogeneous and non-homogeneous solutions?

A homogeneous solution is the general solution to a differential equation where the right-hand side is equal to zero. A non-homogeneous solution is the particular solution to a differential equation where the right-hand side is non-zero.

## 5. Can all differential equations be solved using the method of undetermined coefficients?

No, the method of undetermined coefficients can only be used to solve linear differential equations with constant coefficients and non-homogeneous terms that are polynomials or exponential functions.

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