Particular Solution to a Second Order Ordinary Differential Equation

[aida]

Homework Statement

Find the Particular Solution to the differential Equation satisfying the initial conditions.

y''+7y'+10y= -30
y(0)= 3
y'(0) = -27

Homework Equations

Characteristic Equation for Homogenous Solution
y''+7y'+10y=0

roots are -2 and -5

General Solution is
C1e^-2t + C2e^-5t

The Attempt at a Solution

I figured out the general solution, however I don't know how to go about using the method of undetermined coefficients to solve for the particular solution. Since the g(t) is equal to a constant number and not a function of t.

Thanks for all your help!

 

[mighty2000]

To solve for the particular solution, we can use the method of undetermined coefficients. Since the right-hand side of the differential equation is a constant, we can assume that the particular solution will also be a constant, let's say A.

Substituting this into the differential equation, we get:

A'' + 7A' + 10A = -30

Simplifying, we get:

10A = -30

Therefore, A = -3.

Hence, the particular solution is y = -3.

To find the complete solution, we can add the particular solution to the general solution we found earlier.

Therefore, the complete solution is:

y = C1e^-2t + C2e^-5t - 3

To find the values of C1 and C2, we can use the initial conditions given in the problem.

Using y(0) = 3, we get:

C1 + C2 - 3 = 3

Using y'(0) = -27, we get:

-2C1 - 5C2 = -27

Solving these two equations, we get:

C1 = 12 and C2 = -6

Hence, the particular solution to the given differential equation is:

y = 12e^-2t - 6e^-5t - 3

I hope this helps! Let me know if you have any further questions. Keep up the good work in your studies.