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

by RoRi

February 27, 2016

Let $L$ be defined by

$L(y) = y'' + ay' + by,$

for $a, b \in \mathbb{R}$ constants.

Let $c \in \mathbb{R}$ and $b \neq \omega^2$ . Using the previous exercise prove that the differential equation

$L(y) = c \cos (\omega x)$

has a particular solution given by

$y = A \cos (\omega x - \alpha),$

where

$A = \frac{c}{\sqrt{(b-\omega^2)^2 + a^2 \omega^2}} \quad \text{and} \quad \tan \alpha = \frac{a \omega}{b - \omega^2}.$

Proof. Incomplete.

5 comments

July 13, 2020 at 9:18 am

Anonymous says:

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

In exercise 13 let:

$R(x) = ce^{i\omega x}$

then by exercise 13 we have a solution of $L(y)=R(x)$ given by:

$y=\frac{c}{b-\omega^2 + i\omega a}e^{i\omega x}$

using equation (9.2) to move the imaginary number $b-\omega^2 + i\omega a$ into the numerator we have:

$y=\frac{c(b-\omega^2 - i\omega a)}{(b-\omega^2)^2 + (\omega a)^2}e^{i\omega x}$

expressing the new imaginary number in polar form, where we use $arg(x + iy) = atan(\frac{y}{x})$ derived from the equation above equation (9.5):

$y=\frac{c}{\sqrt{(b-\omega^2)^2 + (\omega a)^2}}e^{i\omega x}e^{i atan{\frac{\omega a}{\omega^2 - b}}}$

invoking Theorem 9.3 we can add together the exponents. Which gives us:

$y= \frac{c}{\sqrt{(b-\omega^2)^2 + (\omega a)^2}}e^{i(\omega x + atan{\frac{\omega a}{\omega^2 - b})}}$

by equation (9.9) we convert to trigonometric form:

$y= \frac{c}{\sqrt{(b-\omega^2)^2 + (\omega a)^2}}\left(\cos ((\omega x + atan\frac{\omega a}{\omega^2 - b})) + i \sin(\omega x + atan\frac{\omega a}{\omega^2 - b})\right)$

using $atan(x) = -atan(-x)$ we arrive at:

$y= \frac{c}{\sqrt{(b-\omega^2)^2 + \omega^2 a^2}}\left(\cos ((\omega x - atan\frac{\omega a}{\omega^2 + b})) + i \sin(\omega x - atan\frac{\omega a}{\omega^2 + b})\right)$

by exercise 12 we know $L(u) = P(x)$ where $u$ is the real part of $y$ and $P(x)$ is the real part of $ce^{i\omega x}$ which is the equation we wanted to prove.

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- July 13, 2020 at 9:28 am
  
  Anonymous says:
  
  In the last expression $\omega^2$ should really be $-\omega^2$ . I apologize for the confusion this might create.
  
  Reply
February 11, 2020 at 11:58 pm

Mike says:

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

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May 31, 2019 at 2:10 pm

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

$\begin{align*} y &= A \cos(\omega x - \alpha) \end{align*}$

is a particular solution to the second order differential equation

$\begin{align*} L(y) &= y'' + ay' + by \\ &= c \cos(\omega x) \end{align*}$

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

$\begin{align*} y &= \frac{A}{2}(e^{i(\omega x - \alpha)} + e^{-i(\omega x - \alpha)}) \\ y' &= \frac{A}{2}(i \omega e^{i(\omega x - \alpha)} - i \omega e^{-i(\omega x - \alpha)}) \\ y'' &= \frac{A}{2}(-\omega^{2} e^{i(\omega x - \alpha)} - \omega^{2} e^{-i(\omega x - \alpha)}) \end{align*}$

L(y) then becomes

$\begin{align*} L(y) &= \frac{A}{2}[(b - \omega^{2})(e^{-i \alpha} e^{i(\omega x)} + e^{i \alpha} e^{-i(\omega x)}) + (a i \omega)(e^{-i \alpha} e^{i(\omega x)} - e^{i \alpha} e^{-i(\omega x)})] \end{align*}$

Rearranging terms gives us

$\begin{align*} L(y) &= \frac{A}{2}[e^{-i \alpha}e^{i \omega x}(b - \omega^{2} + a i \omega) + e^{i \alpha}e^{-i \omega x}(b - \omega^{2} - a i \omega)] \\ &= L(u) + L(v) \end{align*}$

Where

$\begin{align*} u &= \frac{A}{2}e^{-i \alpha}e^{i \omega x} \\ v &= \frac{A}{2}e^{i \alpha}e^{-i \omega x} \end{align*}$

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

$\begin{align*} y &= B e^{i \omega x} \end{align*}$

Is a particular solution to the equation

$\begin{align*} L(y) &= A e^{i \omega x} \end{align*}$

With B defined as follows

$\begin{align*} B &= \frac{A}{b - \omega^{2} + ai \omega} \end{align*}$

And since L(y) is as follows

$\begin{align*} L(y) &= L(u) + L(v) \\ &= \frac{c}{2} e^{i \omega x} + \frac{c}{2} e^{-i \omega x} \end{align*}$

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

$\begin{align*} \frac{A}{2}e^{-i \alpha} &= \frac{c}{2(b - \omega^{2} + ai \omega)} \\ \frac{A}{2}e^{i \alpha} &= \frac{c}{2(b - \omega^{2} - ai \omega)} \end{align*}$

Now, we know that

$\begin{align*} A &= \frac{c}{2 \sqrt{ (b - \omega^{2})^{2} + (a \omega)^{2} }} \end{align*}$

$\begin{align*} \tan(\alpha) &= \frac{a \omega}{b - \omega^{2}} \end{align*}$

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

$\begin{align*} \sin(\alpha) &= \frac{a \omega}{\sqrt{ (b - \omega^{2})^{2} + (a \omega)^{2}}} \\ \cos(\alpha) &= \frac{b - \omega^{2}}{\sqrt{ (b - \omega^{2})^{2} + (a \omega)^{2} }} \end{align*}$

So, we get

$\begin{align*} \frac{A}{2}e^{-i \alpha} &= \frac{c (\cos(\alpha) - i\sin(\alpha))}{2 \sqrt{ (b - \omega^{2})^{2} + (a \omega)^{2} }}\\ &= \frac{c( b - \omega^{2} - ai \omega ) }{2((b - \omega^{2})^{2} + (a \omega)^{2})} \\ &= \frac{c}{2(b - \omega^{2} + ai \omega)} \end{align*}$

$\begin{align*} \frac{A}{2}e^{i \alpha} &= \frac{c (\cos(\alpha) + i\sin(\alpha))}{2 \sqrt{ (b - \omega^{2})^{2} + (a \omega)^{2} }}\\ &= \frac{c( b - \omega^{2} + ai \omega ) }{2((b - \omega^{2})^{2} + (a \omega)^{2})} \\ &= \frac{c}{2(b - \omega^{2} - ai \omega)} \end{align*}$

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).

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May 28, 2019 at 1:12 am

Evangelos says:

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

$\begin{align*} y &= A\cos(\omega x - \alpha) \end{align*}$

and its derivatives

$\begin{align*} y' &= -\omega A\sin(\omega x - \alpha) \\ y'' &= -\omega^{2}A\cos(\omega x - \alpha) \end{align*}$

satisfy the equation

$\begin{align*} L(y) &= y'' + ay' + by \\ &= c\ \cos(\omega x) \end{align*}$

Suppose we have L(y) as follows

$\begin{align*} L(y) &= y'' + ay' + by \\ &= A[(b - \omega^{2}) \cos(\omega x - \alpha) - (a\omega) \sin(\omega x - \alpha) ] \end{align*}$

Using the addition identities for sine and cosine, we get

$\begin{align*} L(y) &= y'' + ay' + by \\ &= A[(b - \omega^{2}) \cos(\omega x - \alpha) - (a\omega) \sin(\omega x - \alpha)] \\ &= A[(b - \omega^{2})(\cos(\omega x) \cos(-\alpha) - \sin(\omega x) \sin(-\alpha)) - (a\omega)(\sin(\omega x) \cos(-\alpha) + \cos(\omega x) \sin(-\alpha))] \\ &= A[(b - \omega^{2}) (\cos(\omega x) \cos(\alpha) + \sin(\omega x) \sin(\alpha)) - (a\omega) (\sin(\omega x) \cos(\alpha) - \cos(\omega x) \sin(\alpha))] \\ &= A[\cos(\omega x) ( (b - \omega^{2}) \cos(\alpha) + (a\omega) \sin(\alpha)) + \sin(\omega x) ( (b - \omega^{2}) \sin(\alpha) - (a\omega) \cos(\alpha) )] \\ \end{align*}$

But if it is the case that $L(y) &= c cos(\omega x)$ , then that means that the factor multiplying $sin(\omega x)$ must be equal to zero.

$\begin{align*} (b - \omega^{2}) \sin(\alpha) - (a\omega) \cos(\alpha) &= 0 \end{align*}$

Or in other words

$\begin{align*} (b - \omega^{2}) \sin(\alpha) &= (a\omega) \cos(\alpha) \end{align*}$

And since $b \neq \omega^{2}$ , we can rearrange terms to get the following

$\begin{align*} \tan(\alpha) &= \frac{a \omega}{ (b - \omega^{2}) } \end{align*}$

But if that is the case, then we also have

$\begin{align*} \tan^{2}(\alpha) &= \frac{(a \omega)^{2}}{ (b - \omega^{2})^{2} } \\ \end{align*}$

Using the trigonometric identity

$\begin{align*} \tan^{2}(\alpha) &= \frac{1}{\cos^{2}(\alpha)} - 1 \\ \end{align*}$

We then get

$\begin{align*} \frac{1}{\cos^{2}(\alpha)} - 1 &= \frac{(a \omega)^{2}}{(b - \omega^{2})^{2}}\\ \end{align*}$

$\begin{align*} \frac{1}{\cos^{2}(\alpha)} &= \frac{ (a \omega)^{2} + (b - \omega^{2})^{2} }{ (b - \omega^{2})^{2} } \\ \end{align*}$

$\begin{align*} \cos^{2} (\alpha) &= \frac{ (b - \omega^{2})^{2} }{ (a \omega)^{2} + (b - \omega^{2})^{2} } \end{align*}$

$\begin{align*} \cos(\alpha) &= \frac{ (b - \omega^{2}) }{ \sqrt{ (a \omega)^{2} + (b - \omega^{2})^{2} } } \end{align*}$

But we also know that

$\begin{align*} \sin^{2}(\alpha) + \cos^{2}(\alpha) &= 1 \end{align*}$

So we get

$\begin{align*} \cos^{2} (\alpha) &= \frac{ (b - \omega^{2})^{2} }{ (a \omega)^{2} + (b - \omega^{2})^{2} } \\ &= 1 - \sin^{2}(\alpha) \end{align*}$

$\begin{align*} \sin^{2} (\alpha) &= 1 - \frac{ (b - \omega^{2})^{2} }{ (a \omega)^{2} + (b - \omega^{2})^{2} }\\ &= \frac{ (a \omega)^{2} }{ (a \omega)^{2} + (b - \omega^{2})^{2} } \end{align*}$

$\begin{align*} \sin (\alpha) &= \frac{ (a \omega) }{ \sqrt{ (a \omega)^{2} + (b - \omega^{2})^{2} } } \end{align*}$

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

$\begin{align*} L(y) &= A \cos(\omega x) [ (b - \omega^{2} ) \cos(\alpha) + (a\omega) \sin(\alpha)) ] \\ &= A \cos(\omega x) \frac{ ( a \omega )^{2} + (b - \omega^{2})^{2} }{ \sqrt{ (a \omega)^{2} + (b - \omega^{2})^{2} } } \\ &= A \sqrt{ (a \omega)^{2} + (b - \omega^{2})^{2} } \cos(\omega x)\\ \end{align*}$

But we know that

$\begin{align*} L(y) &= c\ \cos(\omega x) \end{align*}$

So that must mean that

$\begin{align*} A &= \frac{ c }{ \sqrt{ (a \omega)^{2} + (b - \omega^{2})^{2} } } \end{align*}$

And since we showed earlier that

$\begin{align*} \tan(\alpha) &= \frac{a \omega}{ (b - \omega^{2}) } \end{align*}$

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.

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