Module 4 Homework Solutions

   

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

\[ y'''-6y''+2y'+36y=0 \]

Solution

The characteristic equation, \(\lambda^3-6\lambda^2+2\lambda + 36=0\) has the roots \(\lambda_1=-2\), \(\lambda_2=4+i\sqrt{2}\) , and \(\lambda_3=4-i\sqrt{2}\) . Therefore, the general solution will be

\[ y=c_1e^{-2x}+c_2e^{4+i\sqrt{2}x}+c_3e^{4-i\sqrt{2}x} \]

Problem 5

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 \(\lambda^4-4\lambda^3+7\lambda^2-4\lambda+6=0\) gives us the roots \(\lambda_1=2+i\sqrt{2}\), \(\lambda_2=2-i\sqrt{2}\), \(\lambda_3=i\) and \(\lambda_4=-i\). Thus, giving us the general solution to be

\[ y=c_1e^{2+i\sqrt{2}t}+c_2e^{2-i\sqrt{2}t}+c_3e^{it}+c_4e^{-it} \]

Problem 6

Find the General Solution to a fourth-order linear homogeneous differential equation for with real numbers as coefficients if one solution is known to be

Solution

If \(x^3\)\(e^{4x}\) is a solution, then \(x^2e^{4x}\), \(xe^{4x}\), and \(e^{4x}\) are also solutions. We now hav efour linearly independent solutions to a fourth-order linear, homogeneous differential equation, so we can write the general solution as

\[y(x)=c_4x^3e^{4x}+c_3x^2e^{4x}+c_2xe^{4x}+c_1e^{4x}\]

Problem 7

Construct the differential equation for the previous problem.

Solution

The characteristic equation of the fourth-order differential equation is a fourth-degree polynomial having exactly 4 roots. Because \(x^3 e^{4x}\) is a solution, we know that \(\lambda = 4\) is a root that has a multiplicity of 4. Therefore, the characteristic equation must be \((\lambda - 4)^4 = 0\) or using algebra, we have

\[ \lambda^4 -16\lambda ^3 +96\lambda ^2 - 256\lambda + 256 = 0 \]

Finally, we have the differential equation to be:

\[ y^{(4)} -16y^{(3)} + 96y^{(2)} - 256y^{(1)} + 256 y = 0 \]

Problem 8

Find the General Solution to a third-order linear homogeneous differential equation for \(y(x)\) with real numbers as coefficients if two solutions are known to be and  

Solution

If \(\sin 3x\) is a solution, then so is \(\cos 3x\). Together, with \(e^{-2x}\) we have three linearly independent solutions to a third-order linear, homogeneous differential equation with the general solution being

\[ y(x)=c_1 e^{-2x} + c_2 \cos 3x + c_3 \sin 3x \]

Problem 9

Construct the differential equation for the previous problem.

Solution

The characteristic equation of a third-order differential equation has to have 3 roots. Since \(e^{-2x}\) and \(\sin 3x\) are solutions, then we know that \(\lambda_1 = -2\), \(\lambda_2=3i\) and \(\lambda_3 = -3i\) are the roots of the characteristic equation below

\[ (\lambda+2)(\lambda-3i)(\lambda+3i)=0\\ \lambda ^3 + 2\lambda ^2 + 9\lambda + 18=0 \]

which gives us the solution as

\[ y'''+2y''+9y'+18=0 \]

Problem 10

Solve

\[ y''-y'-2y=4x^2 \]

Solution

Because we have a 2nd order linear homogeneous differential equation, we have the homogeneous solution to be \(y_h=c_1 e^{-x} + c_2 e ^{2x}\) . Also, we have to note that \(\phi(x)=4x^2\) which is a second-degree polynomial. Using Case 1 from Homework 3, we assume our particular solution is

\[ y_p=A_2x^2 + A_1x+A_0 \]

Therefore, \(y_p' = 2A_2x + A_1\) and \(y_p'' = 2A_2\) . When we substitute these values into the differential equation, we have

\[ 2A_2 - \left(2A_2x+A_1\right)-2\left(A_2x^2 +A_1x +A_0\right)=4x^2 \]

Equivalently, we also have

\[ \left(-2A_2\right)x^2 + \left(-2A_2 - 2A_1\right)x+\left(2A_2-A_1-2A_0\right)=4x^2 + 0x + 0 \]

When you equate your coefficients, we have

\[ -2A_2 = 4 \hspace{1cm} -2A_2 - 2A_1 = 0 \hspace{1cm} 2A_2-A_1-2A_0 = 0 \]

When we solve this system of equations, we get

\[ A_2 = -2 \hspace{1cm} A_1 = 2 \hspace{1cm} A_0=-3 \]

Now, we have our particular solution to be

\[ y_p = -2x^2 +2x -3 \]

Putting our homogeneous solution and our particular solution together, we get our general solution, which is

\[ y=y_h+y_p = c_1e^{-x} + c_2e^{2x} - 2x^2+2x-3 \]