Home » Blog » Prove a given equation is a solution of a given differential equation

# Prove a given equation is a solution of a given differential equation

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.

1. Anonymous says:

Very nice proofs by Evangelos. I give my version too.

In exercise 13 let: then by exercise 13 we have a solution of given by: using equation (9.2) to move the imaginary number into the numerator we have: expressing the new imaginary number in polar form, where we use derived from the equation above equation (9.5): invoking Theorem 9.3 we can add together the exponents. Which gives us: by equation (9.9) we convert to trigonometric form: using we arrive at: by exercise 12 we know where is the real part of and is the real part of which is the equation we wanted to prove.

• Anonymous says:

In the last expression should really be . I apologize for the confusion this might create.

2. Mike says:

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

3. Evangelos says:

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).

4. Evangelos says:

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.