Homogeneous, linear ODEs with constant coefficients were likely covered in your Differential Equations course (MTH 256 or equiv.). If you need a review, please see:
Constant Coefficients, Homogeneous
or your differential equations text.
Answer the following questions for each differential equation below:
\(\ddot{x}-\dot{x}-6x=0\)
Order: The highest derivative that appears in this ODE is a second derivative, so the equation is second-order.
Number of Independent Solutions: Thus, there are two linearly independent solutions.
Independent Solutions: I will make the Ansatz \(x(t) = e^{\omega t}\), find possible values of \(\omega\), and then write a general solution that is equal to an arbitrary superposition of the independent solutions. \begin{align*} \omega^{2}e^{\omega t} - \omega e^{\omega t} - 6e^{\omega t} =& 0 \\ \left(\omega - 3\right) \left(\omega + 2\right) =& 0 \end{align*}
The two independent solutions are therefore \(x_{1}= e^{3t}\) and \(x_{2}= e^{-2t}\).
General Solution: \(x = Ae^{3t} + Be^{-2t}\).
Order: The highest derivative that appears in this ODE is a third derivative, so the equation is third order.
Number of Independent Solutions: Thus, there are three linearly independent solutions.
Independent Solutions: I will use the same procedure as in the previous part, starting with the Ansatz \(y(x) = e^{k x}\). \begin{align*} k^{3}e^{kx} - 3k^{2}e^{kx} + 3ke^{kx} - 3e^{kx} =& 0 \\ \left(k-1\right)^{3} = 0 \end{align*}
There is only one solution to this equation, giving \(y_{1}\left(x\right) = e^{x}\). However, I must be able to find two additional (linearly independent) solutions (because this is a third order differential equation). The extra solutions can be found by multiplying the above solution by successively higher powers of x. This gives \(y_{2}\left(x\right) = xe^{x}\) and \(y_{3}\left(x\right) = x^{2}e^{x}\).
General Solution: \(y\left(x\right) = Ae^{x} + Bxe^{x} + Cx^{2}e^{x}\).
Order: The highest derivative that appears in this ODE is a second derivative, so the equation is second order.
Number of Independent Solutions: Thus, there are two linearly independent solutions.
Independent Solutions: I will use the same procedure as in the previous parts, starting with the assumption that \(w\left(z\right) = e^{k z}\). \begin{align*} k^{2}e^{kz}-4ke^{kz} + 5e^{kz} =& 0 \\ k^{2}-4k+5 =& 0 \\ k = 2 \pm i \\ \end{align*}
The last step was done by applying the quadratic formula. The solutions to this equation are \(\omega_{1}\left(z\right) = e^{(2+i)z}\) and \(\omega_{2}\left(z\right) = e^{(2-i)z}\).
Or, you can equivalently write the imaginary exponential part in terms of sines and cosines using Euler relations:
\[\omega_{1}\left(z\right) = e^{2z}(\cos{z}+i\sin{z})\]
\[\omega_{2}\left(z\right) = e^{2z}(\cos{z}-i\sin{z})\].
You can even choose a different set of linearly independent solutions such that one is cosine (and entirely real) and the other is sine (and entirely real) by either adding or subtracting the above solutions (and dividing by a constant):
\[\omega_{+}\left(z\right) = [\omega_{1}\left(z\right)+\omega_{2}\left(z\right)]/2 = e^{2z}\cos{z}\]
\[\omega_{-}\left(z\right) = [\omega_{1}\left(z\right)-\omega_{2}\left(z\right)]/2 = e^{2z}\sin{z}\]
General Solution: Either of these two sets may be used for the superposition that gives the general solution, with different undetermined coefficients: \begin{align} \omega\left(z\right) &= Ae^{(2+i)z}+Be^{(2-i)z}\\ &= Ce^{2z}\sin{z} + De^{2z}\sin{z} \end{align}
Inhomogeneous, linear ODEs with constant coefficients are among the most straigtforward to solve, although the algebra can get messy. This content should have been covered in your Differential Equations course (MTH 256 or equiv.). If you need a review, please see: The Method for Inhomogeneous Equations or your differential equations text.
The general solution of the homogeneous differential equation
\[\ddot{x}-\dot{x}-6 x=0\]
is
\[x(t)=A\, e^{3t}+ B\, e^{-2t}\]
where \(A\) and \(B\) are arbitrary constants that would be determined by the initial conditions of the problem.
Find a particular solution of the inhomogeneous differential equation \(\ddot{x}-\dot{x}-6 x=-25\sin(4 t)\).
I will guess a solution that has the form of the inhomogeneous function, plus all of its derivatives: \(x_{p}\left(t\right) = G\sin(4t) + H\cos(4t)\). Plugging this into the original equation allows us to solve for G and H. \begin{equation*} -16G\sin(4t)-16H\cos(4t)-4G\cos(4t)+4H\sin(4t)-6G\sin(4t)-6H\cos(4t) = -25\sin(4t) \end{equation*} I can then make a sytem of equations by separating the sine and cosine parts: \begin{align*} -16G+4H-6G=-25\\ -16H-4G-6H=0\\ \end{align*} Use your favorite linear algebra techniques to solve this system: \begin{align*} G=\frac{11}{10}\\ H=-\frac{1}{5}\\ \end{align*} Returning to the original assumption gives the particular solution: \begin{equation*} x_{p}(t) = \frac{11}{10}\sin(4t)-\frac{1}{5}\cos(4t) \end{equation*}
Find the general solution of \(\ddot{x}-\dot{x}-6 x=-25\sin(4 t)\).
Once I have the particular solution, I can simply add it to the general solution to the homogeneous equation, which was given in the statment of the problem: \begin{equation*} x_{p}(t) = Ae^{3t} + Be^{-2t} +\frac{11}{10}\sin(4t)-\frac{1}{5}\cos(4t) \end{equation*}
Some terms in your general solution have an undetermined coefficients, while some coefficients are fully determined. Explain what is different about these two cases.
The general solution to the homogeneous equation, which is part of the general solution to the inhomogeneous equation, has two underdetermined coefficients. This is because if you plug any multiple of a solution into the homogeneous equation, the multiplicative factor appears once in every term, and so factors out. On the other hand, the particular solution of the inhomogeneous equation has to have a fixed coefficient. If you try plugging a multiple of this solution into the INHOMOGENEOUS differential equation, the multiplicative factor does NOT appear in the inhomogeneous term. Try it!
Find a particular solution of \(\ddot{x}-\dot{x}-6 x=12 e^{-3 t}\)
I will again guess a solution that has the form of the inhomogeneous function (this time, there are no extra derivative terms to include): \begin{equation*} x_p(t)=Je^{-3t}. \end{equation*} Plugging \(x_p\) into the original equation allows us to solve for \(J\). \begin{align*} 9Je^{-3t}+3Je^{-3t}-6Je^{-3t}&=12e^{-3t}\\ \big(9J+3J-6J\big)e^{-3t}&=12e^{-3t}\\ 6J&=12\\ J&=2. \end{align*} This gives the particular solution \(x_p(t)=2e^{-3t}\). Note that this is a different particular solution that in part (a). When the inhomogeneous term is changed, the particular solution will always change as well.
Find the general solution of \(\ddot{x}-\dot{x}-6 x=12 e^{-3 t}-25\sin(4 t)\)
How is this general solution related to the particular solutions you found in the previous parts of this question?
Can you add these particular solutions together with arbitrary coefficients to get a new particular solution?
I could solve this problem from scratch by another ansatz: the solution will have the form of the inhomogeneous function and its derivatives, but this will lead to redoing all the work I did in previous parts of the problem! When the inhomogeneous function can be split into a sum of two other functions for which the particular solution is known, the particular solution of the full equation is the sum of those individual particular solutions (see section 6.5 of the Linear Algebra book for more detail about this property). Adding together our particular solutions from the previous parts of the problem gives \begin{equation*} x_p(t)=\frac{11}{10}\sin(4t)-\frac{1}{5}\cos(4t)+2e^{-3t}. \end{equation*} Lastly, I add the general solution to the homogeneous equation to get \begin{equation*} x(t)=Ae^{3t}+Be^{-2t}+\frac{11}{10}\sin(4t)-\frac{1}{5}\cos(4t)+2e^{-3t}. \end{equation*} As noted in part (c), you cannot add the particular solutions in any arbitrary combination --- their coefficients were fully determined as part of evaluating the individual particular solutions. I can try this and see what happens with something like \begin{equation*} x_s(t)=M\left[\frac{11}{10}\sin(4t)-\frac{1}{5}\cos(4t)+2e^{-3t}\right]+N\bigg[2e^{-3t}\bigg]. \end{equation*} I can plug \(x_s\) into the left-hand side of \(\ddot{x}-\dot{x}-6x=12e^{-3t}-25\sin(4t)\) and use our results from parts (a) and (d) to simplify to \begin{align*} \ddot{x}_s-\dot{x}_s-6x_s&=12e^{-3t}-25\sin(4t)\\ M\Big[-25\sin(4t)\Big]+N\Big[12e^{-3t}\Big]&=12e^{-3t}-25\sin(4t). \end{align*} Unsurprisingly, this equation is not true for most values of \(M\) and \(N\); only \(M=1\) and \(N=1\) give a valid equation, thus \(x_s\) is only the correct particular solution for those values, and not for arbitrary ones.
Sense-making: Check your answer; Explicitly plug in your final answer in part (e) and check that it satisfies the differential equation.
\begin{align*} 6x(t)&=6Ae^{3t}+6Be^{-2t}+\frac{66}{10}\sin(4t)-\frac{12}{10}\cos(4t)+12e^{-3t} \\ \dot x(t) &= 3Ae^{3t} - 2Be^{-2t} + \frac{44}{10}\cos(4t)+\frac{8}{10}\sin(4t) - 6e^{-3t} \\ \ddot x(t) &= 9Ae^{3t} + 4Be^{-2t} - \frac{176}{10}\sin(4t) + \frac{32}{10}\cos(4t) + 18e^{-3t} \\ \ddot x(t) - \dot x(t) &= 6Ae^{3t} + 6Be^{-2t} - \frac{184}{10}\sin(4t) - \frac{12}{10}\cos(4t) + 24e^{-3t} \\ \ddot{x}-\dot{x}-6 x&=12 e^{-3 t}-25\sin(4 t) \end{align*}
A fun and fascinating fact about sinusoids is that a (normalized) sum of a cosine and a sine function is just a shifted cosine function. You can see this relationship geometrically using the Geogebra applet in https://books.physics.oregonstate.edu/GMM/fourierggb.html. Below is an algebraic proof:
For \(a\) and \(b\) related by the normalization condition \(a^2+b^2=1\), we have \begin{align} a\cos\theta+b\sin\theta &=a\, \frac{e^{i\theta}+e^{-i\theta}}{2} +b\, \frac{e^{i\theta}-e^{-i\theta}}{2i}\\ &=\frac{a-ib}{2}\, e^{i\theta} + \frac{a+ib}{2}\, e^{-i\theta}\\ &=\frac{1}{2} e^{-i\phi}\, e^{i\theta} + \frac{1}{2} e^{i\phi}\, e^{-i\theta}\\ &=\frac{1}{2} e^{i(\theta-\phi)} + \frac{1}{2} e^{-i(\theta-\phi)}\\ &=\cos(\theta-\phi) \end{align} where in (1) we have used Euler's formula, in (2) we have regrouped terms, in (3) we have rewritten \(a+ib=e^{i\phi}\), in (4) we have combined exponents, and in (5) we have used the inverse form of Euler's formulas.
In the rest of this class, we will discuss 4 different forms of the solutions of the equation of motion for a simple harmonic oscillator: \begin{align*} \mbox{A Form: } x(t) &= A\cos(\omega t + \phi) \\[12pt] \mbox{B Form: } x(t) &= B_1\cos(\omega t) + B_2\sin(\omega t)\\[12pt] \mbox{C Form: } x(t) &= Ce^{i\omega t} + C^*e^{-i\omega t}\\[12pt] \mbox{D Form: } x(t) &= Re[De^{i\omega t}] \end{align*}
- A-form: Lines (1-4)
- B-form: Line (1)
- C-Form: Line (5)
For the B-form use a trig identity to reorganize the \(A\) form:
\[\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta\]
\begin{align*} x(t) &= A \cos(\omega_0 t + \phi) \\ &= A\cos(\omega_0 t)\cos\phi -A\sin(\omega_0t)\sin\phi \\ \end{align*} Recognize that the last line looks like the B-form identifying the terms in the last line as \(B_1\) and \(B_2\). \begin{align*} x(t) &= \underbrace{A\cos\phi}_{B_1}\cos(\omega_0 t) +\underbrace{-A\sin\phi}_{B_2}\sin(\omega_0t) \\ \end{align*}
For C-form, rewrite \(\sin\) and \(\cos\) in terms of complex exponentials:
\[\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}\] \[\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\]
\begin{align*} x(t) &= A\cos(\omega_0 t)\cos\phi -A\sin(\omega_0t)\sin\phi \\[12pt] &= A\cos\phi \left[\frac{e^{i\omega_0 t}+e^{-i\omega_0 t}}{2}\right] - A\sin\phi \left[\frac{e^{i\omega_0 t}-e^{-i\omega_0 t}}{2i}\right] \\[12pt] &= A\cos\phi \left[\frac{e^{i\omega_0 t}+e^{-i\omega_0 t}}{2}\right] + iA\sin\phi \left[\frac{e^{i\omega_0 t}-e^{-i\omega_0 t}}{2}\right] \\[12pt] &= \tfrac{A}{2}\left[\cos\phi + i\sin\phi \right] e^{i\omega_0 t} + \tfrac{A}{2}\left[\cos\phi - i\sin\phi \right] e^{-i\omega_0 t} \\[12pt] &= \underbrace{\tfrac{A}{2}e^{i\phi}}_{C}\; e^{i\omega_0 t} + \underbrace{\tfrac{A}{2}e^{-i\phi}}_{C^*}\; e^{-i\omega_0 t} \end{align*}
Therefore, \(|C| = A/2\) (remember, \(A\) is real), and \(\phi\) is the phase of the complex number \(C\).
Alternatively, \(Re(C) = \frac{A}{2} \cos \phi\) and \(Im(C) = \frac{A}{2} \sin \phi\)