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Differential Equations Alpha List
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Differential Equations  Exam A3 
This page contains a complete differential equations exam with worked out solutions to all problems. 
Exam Details  
Tools 
Time  1 hour  
Calculators  no 
Questions  7  
Formula Sheet(s)  none 
Total Points  100  
Other Tools  none 
Instructions:
 This exam is in three main parts, labeled sections 13, with different instructions for each section.
 Show all your work.
 For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
 Correct notation counts (i.e. points will be taken off for incorrect notation).
 Give exact answers.
Find the general solution for each of the following differential equations. Each question in this section is worth 5 points.
Question 1 
\( y^{(3)}4y' = 0 \)



\( y = A + Be^{2t} + Ce^{2t} \)

\( r^3  4r = r(r^24) = 0 \) \( \to \) \( r=0 ~~~ r^24=0 \) \( \to \) \( r^2 = 4 ~~~ r = \pm 2 \)
\( y = A + Be^{2t} + Ce^{2t} \)
Question 1 Final Answer 
\( y = A + Be^{2t} + Ce^{2t} \)

Question 2 
\( y^{(4)} + 2y''+y=0 \)



\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \)

\(
\begin{array}{rcl}
r^4 + 2r^2 + 1 & = & 0 \\
(r^2+1)^2 & = & 0 \\
r^2 + 1 & = & 0 ~~~ \text{multiplicity 2} \\
r^2 & = & 1 \\
r = \pm i
\end{array}
\)
\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \)
Question 2 Final Answer 
\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \)

Question 3 
\( y^{(3)}3y''+3y'y = 0 \)



\( y = (A+Bt+Ct^2)e^t \)

\(
\begin{array}{rcl}
r^3  3r^2 + 3r  1 & = & 0 \\
(r1)(r^22r+1) & = & 0 \\
(r1)(r1)^2 & = & 0 \\
(r1)^3 & = & 0 \\
r1 & = & 0 ~~~ \text{multiplicity 3} \\
r & = & 1
\end{array}
\)
\( y = (A+Bt+Ct^2)e^t \)
Question 3 Final Answer 
\( y = (A+Bt+Ct^2)e^t \)

Find the general solution for each of the following differential equations. Use the method of undetermined coefficients to find the particular solutions. Each question in this section is worth 15 points.
Question 4 
\( y^{(3)} 4y' = 4e^{2t} \)



\( \displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{2t} + \frac{te^{2t}}{2} }\)

First, we solve the homogeneous equation \( y^{(3)} 4y' = 0 \).
\(
\begin{array}{rcl}
r^3  4r & = & 0 \\
r(r^24) & = & 0 \\
r = 0 & & r^2 = 4 ~~~ \to ~~~ r = \pm 2
\end{array}
\)
\( y_h = A + Be^{2t} + Ce^{2t} \)
Now we need to find the particular solution associated with \( 4e^{2t} \).
Since we already have an \( e^{2t} \) factor in the homogeneous solution, we need to use \( y_p = Dte^{2t} \), which includes the extra factor of \( t \) to get an independent solution.
\(
\begin{array}{rcl}
y'_p & = & Dt(2e^{2t}) + De^{2t} = (2Dt+D)e^{2t} \\
y''_p & = & (2Dt+D)e^{2t}(2) + (2D)e^{2t} = (4Dt+4D)e^{2t} \\
y^{(3)}_p & = & (4Dt+4D)e^{2t}(2) + e^{2t}(4D) = (8Dt+12D)e^{2t}
\end{array}
\)
\(
\begin{array}{rcl}
y^{(3)} 4y' & = & 4e^{2t} \\
(8Dt+12D)e^{2t}  4(2DT+D)e^{2t} & = & 4e^{2t} \\
(12D  4D)e^{2t} & = & 4e^{2t} \\
8D & = & 4 \\
D & = & 1/2
\end{array}
\)
\( \displaystyle{ y_p = \frac{te^{2t}}{2} }\)
Combining the homogeneous and particular solutions gives us the general solution.
Question 4 Final Answer 
\( \displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{2t} + \frac{te^{2t}}{2} }\)

Question 5 
\( y^{(3)}4y' = 5\cos(t) \)



\(y_p=\sin(t)\)

We found the homogeneous solution in the last problem. So here we need to find the particular solution. Looking at the right side of the equation, the form of our solution is \(y_p = A\cos(t)+B\sin(t)\).
\(y_p = A\cos(t)+B\sin(t)\) 
\(y'_p=A\sin(t)+B\cos(t)\) 
\(y''_p=A\cos(t)B\sin(t)\) 
\(y^{(3)}_p=A\cos(t)B\cos(t)\) 
Now plug in the derivatives above into the original differential equation to solve for A and B.
\(y^{(3)}4y' = (A\cos(t)B\cos(t))4(A\sin(t)+B\cos(t)) = \) \(5A\sin(t)5B\cos(t) = 5\cos(t))\) \(\to A=0, B=1\)
Question 5 Final Answer 
\(y_p=\sin(t)\)

Solve the following problems.
Question 6 
[15 points] Solve the initial value problem \( y^{(4)} + 13y''+36y = 0; ~~~ y(0)=5, y'(0)=y''(0) = y^{(3)}= 0 \)



\(y=9\cos(2t)4\cos(3t)\)

For this solution, we can apply the techniques of second order linear equations with constant coefficients. So our characteristic polynomial is \(r^4+13r^2+36=0\) \(\to\) \((r^2+4)(r^2+9)=0\) \(\to\) \(r=\pm2i,\pm3i\).
This gives us the general form of the solution as \(y=A\cos(2t)+B\sin(2t)+C\cos(3t)+D\sin(3t)\).
Taking derivatives and solving for the coefficients, gives us \(A=9, B=0, C=4, D=0\).
Question 6 Final Answer 
\(y=9\cos(2t)4\cos(3t)\)

Question 7 
[40 points] Find the general solution and use the method of variation of parameters to find the particular solution for \( y^{(3)}  3y''+3y'y=6e^t \).



\(y=Ae^t+Bte^t+Ct^2e^t+t^3e^t\)

This is a rather long problem but the 3x3 determinants reduce to manageable equations. We will show some intermediate results without all the details for calculating determinants.
For the homogeneous solution, we have constant coefficents, so our characteristic equation is \(r^33r^2+3r1=0\) \(\to\) \((r1)^3=0\). This gives us the homogeneous solution \(y_h=Ae^t+Bte^t+Ct^2e^t\).
Now we need to set up the determinants for the variation of parameters method.
\(y_1=e^t\)  \(y_2=te^t\)  \(y_3=t^2e^t\) 
The Wronskian is
\(W = e^{3t}
\begin{vmatrix}
1 & t & t^2 \\
1 & t+1 & t^2+2t \\
1 & t+2 & t^2+4t+2
\end{vmatrix}
\)
This looks messy but it reduces to \(W=2e^{3t}\). The variation of parameters technique gives us the following results.
\(u'_1=3t^2\) \(\to\) \(u_1=t^3\) 
\(u'_2=6t\) \(\to\) \(u_2=3t^2\) 
\(u'_3=3\) \(\to\) \(u_3=3t\) 
To get the particular solution, we use these results in the equation \(y_p=u_1y_1+u_2y_2+u_3y_3=t^3e^t\) and our answer is \(y=y_h+y_p\).
Question 7 Final Answer 
\(y=Ae^t+Bte^t+Ct^2e^t+t^3e^t\)
