Let be defined by

for constants.

Let and . Using the previous exercise prove that the differential equation

has a particular solution given by

where

*Proof.* **Incomplete.**

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Stumbling Robot

A Fraction of a Dot
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Prove a given equation is a solution of a given differential equation

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Let be defined by

for constants.

Let and . Using the previous exercise prove that the differential equation

has a particular solution given by

where

*Proof.* **Incomplete.**

By application of the results of exercises 12 and 13 one can get relatively quickly to the solution of this exercise.

I’ll do a second version of the answer, since the exercise asked to use the results of exercise 13, which I didn’t do the first time around. So…

We want to show that

is a particular solution to the second order differential equation

Now, assuming x is real, we can use the definition of cosine from exercise 4 to re-write our equation for y, as well as its derivatives, as follows

L(y) then becomes

Rearranging terms gives us

Where

But, from the results of exercise 13, we know that for complex A and real omega, with b not equal to omega squared, there exists a constant B such that

Is a particular solution to the equation

With B defined as follows

And since L(y) is as follows

We wish to show that u and v are solutions to L(u) and L(v). If this is true, then it must be that

Now, we know that

And, as computed below, we have sine and cosine defined as follows

So, we get

Thus, we have shown that u and v are particular solutions to L(u) and L(v), and by extension, that y = u + v is a particular solution to L(y) = L(u) + L(v).

We wish to show that there exists a constant A, and an angle alpha such that the equation

and its derivatives

satisfy the equation

Suppose we have L(y) as follows

Using the addition identities for sine and cosine, we get

But if it is the case that , then that means that the factor multiplying must be equal to zero.

Or in other words

And since , we can rearrange terms to get the following

But if that is the case, then we also have

Using the trigonometric identity

We then get

But we also know that

So we get

So, since we have the sine and cosine of the angle alpha in terms of a, b, and omega, we can re-write L(y) as follows

But we know that

So that must mean that

And since we showed earlier that

Thus, we have shown that there exists a constant A and an angle alpha that satisfy the differential equation L(y). This completes the proof.