Module 3 Homework Solutions
James Boffenmyer
03/03/26
Below are the solutions to the Module 3 Homework Assignment for Second-Order Differential Equations. Prior to these solutions, let’s recall a few important things that I’ll have to refer back to periodically.
An n-th Order Linear ODE
An n-th order Linear Differential Equation will always have the form of
\[ b_n(x)y^{(n)}+b_{n-1}(x)y^{(n-1)} + \cdots + b_2(x)y^{''}+b_1(x)y^{'} + b_0(x)y=g(x) \]
This is for 1st order, 2nd order, 3rd order, and so on. In this case, \(g(x)\) and the coefficients \(b_j(x)\) where \(j=0,1,2,\ldots, n\) depend solely on \(x\). In other words, they do not depend whatsoever on \(y\) or and derivative of \(y\).
Homogeneous vs Non-Homogeneous
If then the above equation is Homogeneous. If it's not, then it's Non-Homogeneous.
General Solutions
Once you derive your characteristic equation, then you have three possible cases for your General Solution
Case 1: \(\lambda_1\) and \(\lambda_2\) are Real & Distinct
Two linearly independent solutions are \(e^{\lambda_1x}\) and \(e^{\lambda_2x}\) and the general solution is given by
\[ y=c_1 e^{\lambda_1x} +c_2e^{\lambda_2x} \]
Special Case
In the special case that , the solution above can be written as
Case 2: When \(\lambda_1\) & \(\lambda_2\) are Complex
Since \(a_1\) and \(a_0\) are assumed to be real in your characteristic equation, the roots of the characteristic equation must appear in conjugate pairs. Therefore, the other root is \(\lambda_2=a-bi\) . Two linearly independent solutions are \(e^{(a+bi)x}\) and \(e^{(a-bi)x}\) and the general complex solution would be
\[y=d_1e^{(a+bi)x} + d_2e^{(a-bi)x} \]
which, is equivalent to a different form that some of us are used to seeing
\[ y=c_1e^{ax} \cos bx + c_2e^{ax} \sin bx \]
Case 3: When \(\lambda_1=\lambda_2\)
Two linearly independent solutions are \(e^{\lambda_1x}\) and \(xe^{\lambda_1x}\) and the general solution is given by
\[ y=c_1e^{\lambda_1x}+c_2xe^{\lambda_1x} \]
Warning
These cases are not valid if the DE is not linear or does not have constant coefficients.
Problem 1
Solve \[\ddot{x}-0.01x=0\]
Solution
The characteristic equation is
\[ \lambda^2-0.01=0 \]
This equation can be factored into
\[ (\lambda-0.1)(\lambda+0.1)=0 \]
By using Algebra, the roots for this equation are \(\lambda_1=0.1\) and \(\lambda_2=-0.1\). Note, they are real numbers and distinct. Therefore, the general solution to this equation would be
\[ y=c_1e^{0.1t}+c_2e^{-0.1t} \]
or, if you want to use the special case version, we have:
\[ y=c_1 \cosh 0.1t + c_2 \sinh 0.1t \]
Problem 2
Solve
\[ y''+4y'+5y=0 \]
Solution
The characteristic equation is
\[ \lambda^2+4\lambda+5=0 \]
Once we break out the good ’ole quadratic formula,
\[ \lambda = \frac{-b\pm \sqrt{b^2-4ac}}{2a} \]
We have
\[ \lambda = \frac{-4\pm \sqrt{4^2-4(1)(5)}}{2(1)} \]
which gives us the following roots \(-2 +i\) and \(-2-i\) or simply \(-2 \pm i\)
Remember from algebra, complex solutions are written in the form \(a\pm bi\) ? In this case, \(a=-2\) and \(b=1\) . Therefore, the general solution, because we have a complex pair of roots, is
\[ y=k_1e^{-2x}\cos x + k_2e^{-2x}\sin x \]
Problem 3
Solve
\[ y^{''}-3y'+4y=0 \]
Solution
First, find the characteristic equation, which is
\[ \lambda^2-3\lambda +4=0 \]
Using the quadratic formula to solve this quadratic equation, we get the following roots:
\[ \lambda = \frac{-(-3)\pm \sqrt{(-3)^2-4(1)(4)}}{2(1)} = \frac{3}{2}\pm \frac{\sqrt{7}}{2}i \]
Since \(a=\frac{3}{2}\) and \(b=\frac{\sqrt{7}}{2}\) and by using Case 2 above, we’ll have our general solution as:
\[ y=c_1e^{\frac{3}{2}x} \cos \frac{\sqrt{7}}{2} x + c_2e^{\frac{3}{2}x}\sin \frac{\sqrt{7}}{2}x \]
Problem 4
Solve
\[ \ddot{y}-6\dot{y}+25y=0 \]
Solution
First, the characteristic equation of this would be
\[ \lambda^2-6\lambda + 25 = 0 \]
Using the Quadratic Formula, yields our roots, which are
\[ \lambda = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(25)}}{2(1)} = 3\pm 4i \]
Since we have complex roots, our \(a=3\) and \(b=4\), giving our general solution
\[ y=c_1e^{3t} \cos 4t + c_2e^{3t}\sin 4t \]
Problem 5
Solve
\[ \frac{d^2I}{dt^2}+20\frac{dI}{dt}+200I = 0 \]
Solution
First, find the characteristic equation. Even though it’s written in derivative form, we can still build our characteristic equation. Thus, we’ll have
\[ \lambda^2+20\lambda + 200=0 \]
Yet again, using the Quadratic formula to solve for \(\lambda\), we’ll get our roots, which are
\[ \lambda = \frac{-20\pm \sqrt{20^2-4(1)(200)}}{2(1)} = -10\pm 10i \]
Since we have a complex roots, our \(a = -10\) and \(b=10\). Therefore, our general solution would be
\[ I=c_1e^{-10t} \cos 10t + c_2e^{-10t}\sin 10t \]
Problem 6
Solve
\[ y^{(4)}-9y^{''}+20y=0 \]
Firstly, the characteristic equation would be
\[ \lambda^4-9\lambda + 20=0 \]
Using long division, synthetic division, or factoring (whichever you prefer), we get our roots to be
\[ x=\pm \sqrt{5} \text{ and } x = \pm 2 \]
Therefore, we have our general solution to be either one of the two below:
\[ \begin{eqnarray*} y&=&c_1e^{2x}+c_2e^{-2x}+c_3e^{\sqrt{5}x}+c_4e^{-\sqrt{5}x}\\ &&\\ y&=& k_1\cosh 2x + k_2 \sinh{2x}+k_3\cosh \sqrt{5}x + k_4\sinh \sqrt{5}x \end{eqnarray*} \]
Problem 7
Solve
\[ \frac{d^4x}{dt^4}-4\frac{d^3x}{dt^3}+7\frac{d^2x}{dt^2}-4\frac{dx}{dt}+6x=0 \]
Solution
The characteristic equation would be
\[ \lambda^4-4\lambda^3+7\lambda^2-4\lambda+6=0 \]
Using Synthetic Division to work this problem down into quadratic form, then using the quadratic formula, we obtain the roots
\[ \lambda_1=2+\sqrt{2}i, \hspace{3mm} \lambda_2=2-\sqrt{2}i, \hspace{3mm} \lambda_3=i, \hspace{3mm} \lambda_4=-i \]
Therefore, since we have complex roots, we have our general solution to be
\[ x=d_1e^{(2+\sqrt{2}i)t}+d_2e^{(2-\sqrt{2}i)t}+d_3e^{it}+d_4e^{-it} \]