Complex and Resonant Solutions for First and Second Order Equations
Return to the Math 2250 Overview Page.
Euler's Formula and Complex Roots
Many differential equations can be simplified using Euler's Formula (Also called Euler's Identity). This formula can be derived from Taylor series approximations of Sine, Cosine, and exponential functions. These approximations are \[f(t)=\sum_{n=0}^{\infty}\frac{f^n(a)}{n!}(t-a)\]\[e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+...\]\[\sin(t)=t-\frac{t^3}{6}+\frac{t^5}{120}+...\]\[\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+...\]
By inspecting the Taylor series expansions of all three terms, a similarity is apparent. If we add the sine and cosine expansions, they are close to equivalent to the exponential function. An oscillating pattern from positive to negative terms can be observed. To make these terms equivalent, we will multiply the sine term by i and raise e to the power of it. The Taylor series expansion of these equations now becomes
\[e^{it}=1+it-\frac{t^2}{2}-\frac{it^3}{6}+\frac{t^4}{24}+\frac{it^5}{120}+...\]\[i\sin(t)=it-\frac{it^3}{6}+\frac{it^5}{120}+...\]
By adding the \(i\sin(t)\) and \(\cos(t)\) expansions, we find that they are exactly equivalent to \(e^{it}\) expansion. The complete expression of this equality is known as Euler's formula and is written as\[e^{it}=\cos(t)+i\sin(t)\]
or\[Re^{i\omega t}=R\cos(\omega t)+iR\sin(\omega t)\]
Deriving Trigonometric Identities using Euler's Formula
Euler's Formula substitutions can be used to solve for various common trigonometric identities. This section will cover an example of solving for the double-angle formulas of Sine and Cosine simultaneously. First, write the Euler's Formula substitution for \(2\omega\).\[e^{2i\omega}=\cos(2\omega)+i\sin(2\omega)\]
Rewrite the same equality by decomposing the complex exponential\[e^{i\omega}e^{i\omega}=(\cos(\omega)+i\sin(\omega))(\cos(\omega)+i\sin(\omega))\]
Distribute the terms in the second equality and set it equal to the first\[\cos^2(\omega)+2i\sin(\omega)\cos(\omega)-\sin^2(\omega)=\cos(2\omega)+i\sin(2\omega)\]
Solve for the real and imaginary components individually\[\cos^2(\omega)-\sin^2(\omega)=\cos(2\omega)\]\[2\sin(\omega)\cos(\omega)=\sin(2\omega)\]
These are variations of the double-angle identities. The first can be further adapted using a Pythagorean identity\[\cos^2(\omega)-(1-\cos^2(\omega))=\cos(2\omega)\]\[2\cos^2(\omega)=1+\cos(2\omega)\]\[\cos^2(\omega)=\frac{1+\cos(2\omega)}{2}\]
Or\[\sin^2(\omega)=\frac{1-\sin(2\omega)}{2}\]
Solutions to Differential Equations using Euler's Formula
Euler's identity was briefly introduced to explain the behavior of certain second-order differential equations. Euler's formula can also be used as a powerful tool in solving first-order differential equations as well as solving sinusoidal forcing functions. Let's review two potential differential equations.\[x' - 7x = 3\cos(2t)\]\[x(0)=10\]
This first-order differential equation can be solved using an Euler's formula substitution. That substitution can be written as\[x' - 7x = Re[3e^{2it}]\]
Solve by using integrating factor\[(xe^{-7t})'=3e^{(-7+2i)t}\]\[\int_0^t(xe^{-7t})'dt=\int_0^t3e^{(-7+2i)t}dt\]\[xe^{-7t}-10=\frac{3}{-7+2i}e^{(-7+2i)t}-\frac{3}{-7+2i}\]
Rationalize the complex term using a complex conjugate\[\frac{3}{-7+2i}(\frac{-7-2i}{-7-2i})=\frac{-21-6i}{-44}\]\[xe^{-7t}=10+\frac{21+6i}{44}e^{(-7+2i)t}-\frac{21+6i}{44}\]
Divide by exponential term on \(\textit{LHS}\)\[x(t)=10e^{7t}+\frac{21+6i}{44}e^{(2i)t}-\frac{21+6i}{44}e^{7t}\]
Reintroduce Sine and Cosine using Euler's formula. As the original forcing term only had the Cosine term, we will use only the real parts of the solution in the complete solution
\[x_p=Re[(\frac{21+6i}{44})(\cos(2t)+i\sin(2t))-\frac{21+6i}{44}e^{7t}]=\frac{21}{44}\cos(2t)-\frac{3}{22}\sin(2t)-\frac{21}{44}e^{7t}\]\[x(t)=10e^{7t}+\frac{21}{44}\cos(2t)-\frac{3}{22}\sin(2t)-\frac{21}{44}e^{7t}\]
or\[x(t)=\frac{419}{44}e^{7t}+\frac{21}{44}\cos(2t)-\frac{3}{22}\sin(2t)\]
Resonance Solutions
Resonance occurs when the rates associated with the homogeneous differential equation match the rate of the forcing function. This will cause general solutions to no longer satisfy the original differential equation. When this situation is encountered, the solution can be found by using Green's convolution with an appropriate impulse response. Consider the following example.\[x''+ 2x'+ x= e^{-t}\]
The homogeneous solution to this differential is a critically damped harmonic oscillator as shown by the characteristic polynomial.\[r^2+2r+r=0\]\[r=-1\]
The general solution can be written as\[C_1e^{-t}+C_2te^{-t}\]
The root of the homogeneous solution matches that of the forcing function. Attempting to solve this problem with the method of constant coefficients would result in a \(0=C\) scenario. To solve for the particular solution, we will apply the impulse response for a critically damped system and solve using Green's function.
\[g(t)=te^{-t}\]\[x_p=\int_0^t (t-s)e^{-(t-s)}e^{-s}ds\]\[e^{-t}\int_0^t (t-s)ds\]\[e^{-t}\bigg(t^2-\frac{t^2}{2}\bigg)\]\[x_p=\frac{t^2}{2}e^{-t}\]
Our complete solution to this differential equation can now be assembled from the particular and null solution
\[x(t)=C_1e^{-t}+C_2te^{-t}+\frac{t^2}{2}e^{-t}\]
This method can be used to solve any resonant linear differential equation. The general form of the impulse response for a second-order differential equation can be found in the explanation of Green's Function.