Undetermined Coefficients - Differential Equations

This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us .

This tells us that the homogeneous solution is . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that . Then,

and

Plugging into the original equation, we have

Which implies that and . Solving via substitution,

Thus, the particular solution is and the overall solution is the particular plus the homogeneous.

So

We first note that the differential equation has characteristic equation

,

since the roots of this characteristic polynomial are linearly independent of the forcing function

,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form

We first figure that the forcing function is linearly independent to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

We start with the assumption that the particular solution must be of the form

.

Then we solve the first and second derivatives with this assumption, that is,

and .

Then we plug in these quantities into the given equation to get:

, which solves for .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

The form of a guess for a particular solution is

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us .

This tells us that the homogeneous solution is . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that . Then,

and

Plugging into the original equation, we have

Which implies that and . Solving via substitution,

Thus, the particular solution is and the overall solution is the particular plus the homogeneous.

So

We first note that the differential equation has characteristic equation

,

since the roots of this characteristic polynomial are linearly independent of the forcing function

,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form

We first figure that the forcing function is linearly independent to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

We start with the assumption that the particular solution must be of the form

.

Then we solve the first and second derivatives with this assumption, that is,

and .

Then we plug in these quantities into the given equation to get:

, which solves for .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

The form of a guess for a particular solution is

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us .

This tells us that the homogeneous solution is . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that . Then,

and

Plugging into the original equation, we have

Which implies that and . Solving via substitution,

Thus, the particular solution is and the overall solution is the particular plus the homogeneous.

So

We first note that the differential equation has characteristic equation

,

since the roots of this characteristic polynomial are linearly independent of the forcing function

,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form

We first figure that the forcing function is linearly independent to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

We start with the assumption that the particular solution must be of the form

.

Then we solve the first and second derivatives with this assumption, that is,

and .

Then we plug in these quantities into the given equation to get:

, which solves for .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.