logo    

Below are the solutions to the Module 4 Homework Assignment for Higher-Order Differential Equations. Prior to these solutions, let’s recall a few important things that I’ll have to refer back to periodically.

The Characteristic Equation

An n-th order Linear Differential Equation will always have the form of

\[ \begin{equation} y^{(n)}+a_{n-1}y^{(n-1)}+\cdots + a_1y'+a_0y=0 \end{equation} \]

with constant coefficients \(a_j (j=0,1,2,\ldots,n-1)\) is

\[ \lambda^n+a_{n-1}\lambda^{n-1}+\cdots+ a_1\lambda^{1}+a_o\lambda^{0}=0 \]

Keep in mind, the second equation is obtained from the first equation by replacing \(y^j\) by \(\lambda^j (j=0,1,2,\ldots,n-1)\). Characteristic equations for DEs having dependent variables other than \(y\) are obtained analogously, by replacing the \(j^{th}\) derivative of the dependent variable by \(\lambda^j (j=0,1,2,\ldots, n-1)\) .

 

The General Solution

Once you derive your characteristic equation, then you can derive your General Solution

The roots of the characteristic equation determine the solution of the first equation above. If the roots \(\lambda_1,\lambda_2, \ldots ,\lambda_n\) are all real and distinct, then the solution would be

\[ y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}+\cdots+c_ne^{\lambda_nx} \]

 

Problem 1

Solve \[ y'''-6y''+11y'+6y=0 \]

Solution

The characteristic equation is

\[ \lambda^3-6\lambda^2+11\lambda-6=0 \]

By using synthetic division, we can factor it into the following form

\[ (\lambda - 1)(\lambda-2)(\lambda-3)=0 \]

Then the roots would be \(\lambda_1=1\), \(\lambda_2 = 2\) and \(\lambda_3=3\)

Now that we have that, we can derive our general solution, which is

\[ y=c_1e^{1x}+c_2e^{2x}+c_3e^{3x} \]

Problem 2

Solve

\[ y^{(4)}-9y''+20y=0 \]

Solution

The characteristic equation is

\[ \lambda^4-9\lambda^2+20=0 \]

which can be factored into

\[ (\lambda-2)(\lambda+2)(\lambda-\sqrt{5})(\lambda+\sqrt{5})=0 \]

Thus, indicating our roots would be \(\lambda_1=2\), \(\lambda_2=-2\) , \(\lambda_3 = \sqrt{5}\) and \(\lambda_4=-\sqrt{5}\)

Therefore, our final general solution would be:

\[ y=c_1e^{2x}+c_2e^{-2x}+c_3e^{\sqrt{5}x}+c_4e^{-\sqrt{5}x} \]

You can also write it using hyperbolic trigonometric function notation, which in this case it would be

\[ y=k_1\cosh (2x)+k_2\sinh(2x)+k_3\cosh(\sqrt{5}x) + k_4\sinh(\sqrt{5}x) \]

Problem 3

Solve

\[ y'-5y=0 \]

Solution

First, find the characteristic equation, which is

\[ \lambda-5=0 \]

giving us the root \(\lambda_1=5\)

Since it’s only one root, our general solution would be

\[ y=c_1e^{5x} \]

Problem 4

Solve

Solution

Problem 5

Solve

Solution

Problem 6

Solve

Solution

Problem 7

Solve

Solution

Problem 8

Solve

Solution

Problem 9

Solve

Solution

Problem 10

Solve

Solution