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

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.

2 comments

  1. 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).

  2. 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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