Construct DE w/ Soln y(t) = e^t cos(3t)

[LocalStudent]

Hi

I'm working through some example question and the memo leaves out this question. I was hoping someone can help me out.

The question:
Construct a differential equation of the form y'' + by' + cy = 0 which has y(t) = e^t cos(3t) as one of its solutions.

What I did:
First I found the y'(t) and y''(t)
Plugged that into the DE.

My answer:
b = -2
c = -6

y'' - 2y' - 6y = 0

 

[Simon Bridge]

Check your answer by plugging the solution into it from scratch.
It will also help to show your working step-by-step.

 

[mfb]

You can use WolframAlpha to check your result.

 

[pasmith]

Hi

I'm working through some example question and the memo leaves out this question. I was hoping someone can help me out.

The question:
Construct a differential equation of the form y'' + by' + cy = 0 which has y(t) = e^t cos(3t) as one of its solutions.

What I did:
First I found the y'(t) and y''(t)
Plugged that into the DE.

My answer:
b = -2
c = -6

y'' - 2y' - 6y = 0

Hint: What is the relationship between the ODE
$$y'' + by' + cy = 0$$
and the quadratic equation
$$\lambda^2 + b\lambda + c = 0?$$

You may also find the identity
$$\cos kt \equiv \frac{e^{ikt} + e^{-ikt}}{2}$$
helpful.

 

[blue_raver22]

Solution:
To construct a differential equation with the given solution, we can use the method of undetermined coefficients. We know that the solution is of the form y(t) = e^t cos(3t), so we can assume that the differential equation will also have a term of e^t cos(3t).

Using the product rule, we can find the derivatives of y(t):
y'(t) = e^t (-sin(3t) + 3cos(3t))
y''(t) = e^t (-4sin(3t) + 9cos(3t))

Now, we can substitute these into the differential equation:
y'' + by' + cy = 0
e^t (-4sin(3t) + 9cos(3t)) + b(e^t (-sin(3t) + 3cos(3t))) + c(e^t cos(3t)) = 0

To simplify, we can group the terms with the same trigonometric functions:
e^t (9cos(3t) + 3bcos(3t) + ccos(3t)) + e^t (-4sin(3t) - bsin(3t)) = 0

Now, we can equate the coefficients of e^t cos(3t) and e^t sin(3t) to get our values for b and c:
9 + 3b + c = 0
-4 - b = 0

Solving for b and c, we get b = -4 and c = -6. Therefore, our differential equation is:
y'' - 4y' - 6y = 0

This satisfies the given solution of y(t) = e^t cos(3t) and is a valid differential equation.