This page contains a complete differential equations exam with worked out solutions to all problems.
Note  This exam is not from a specific semester. See the list of topics that it covers to determine if it applies to you.
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Exam Details  

Time  1 hour 
Questions  7 
Total Points  100 
Tools  

Calculator  no 
Formula Sheet(s)  no 
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, simplified answers.
Section 1
Find the general solution for each of the following differential equations. Each question in this section is worth 5 points.
\( y^{(3)}4y' = 0 \)
Problem Statement 

Find the general solution to \( y^{(3)}4y' = 0 \)
Final Answer 

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

Find the general solution to \( y^{(3)}4y' = 0 \)
Solution 

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

\( y = A + Be^{2t} + Ce^{2t} \)
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\( y^{(4)} + 2y'' + y = 0 \)
Problem Statement 

Find the general solution to \( y^{(4)} + 2y'' + y = 0 \)
Final Answer 

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

Find the general solution to \( y^{(4)} + 2y'' + y = 0 \)
Solution 

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

\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \)
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\( y^{(3)}  3y''+3y'  y = 0 \)
Problem Statement 

Find the general solution to \( y^{(3)}  3y''+3y'  y = 0 \)
Final Answer 

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

Find the general solution to \( y^{(3)}  3y''+3y'  y = 0 \)
Solution 

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

\( y = (A+Bt+Ct^2)e^t \)
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Section 2
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.
\( y^{(3)} 4y' = 4e^{2t} \)
Problem Statement 

Use the method of undetermined coefficients to find the general solution to \( y^{(3)} 4y' = 4e^{2t} \)
Final Answer 

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

Use the method of undetermined coefficients to find the general solution to \( y^{(3)} 4y' = 4e^{2t} \)
Solution 

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 obtain an independent solution.
\( y'_p = Dt(2e^{2t}) + De^{2t} = (2Dt+D)e^{2t} \) 
\( y''_p = (2Dt+D)e^{2t}(2) + (2D)e^{2t} = (4Dt+4D)e^{2t} \) 
\( y^{(3)}_p = (4Dt+4D)e^{2t}(2) + e^{2t}(4D) = (8Dt+12D)e^{2t} \) 

\( y^{(3)} 4y' = 4e^{2t} \) 
\( (8Dt+12D)e^{2t}  4(2DT+D)e^{2t} = 4e^{2t} \) 
\( (12D  4D)e^{2t} = 4e^{2t} \) 
\( 8D = 4 \) 
\( D = 1/2 \) 

\( \displaystyle{ y_p = \frac{te^{2t}}{2} }\) 
Combining the homogeneous and particular solutions gives us the general solution.
Final Answer 

\( \displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{2t} + \frac{te^{2t}}{2} }\)
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\( y^{(3)}4y' = 5\cos(t) \)
Problem Statement 

Use the method of undetermined coefficients to find the general solution to \( y^{(3)}4y' = 5\cos(t) \)
Final Answer 

\( y_p = \sin(t) \)
Problem Statement 

Use the method of undetermined coefficients to find the general solution to \( y^{(3)}4y' = 5\cos(t) \)
Solution 

We found the homogeneous solution in the previous 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.
\(y^{(3)}4y' = (A\cos(t)B\cos(t))4(A\sin(t)+B\cos(t)) = \) \(5A\sin(t)5B\cos(t) = 5\cos(t))\) \(\to A=0, B=1\)
Final Answer 

\( y_p = \sin(t) \)
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Section 3
Solve the following problems.
Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Problem Statement 

Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Final Answer 

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

Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Solution 

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 \( A=9, B=0, C=4, D=0 \).
Final Answer 

\( y = 9\cos(2t)  4\cos(3t) \)
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Find the general solution and use the method of variation of parameters to find the particular solution for \( y^{(3)}  3y''+3y'  y = 6e^t \).
Problem Statement 

Find the general solution and use the method of variation of parameters to find the particular solution for \( y^{(3)}  3y''+3y'  y = 6e^t \).
Final Answer 

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

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

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 \( y_p = u_1 y_1 + u_2 y_2 + u_3 y_3 = t^3e^t \) and our answer is \( y = y_h + y_p \).
Final Answer 

\( y = Ae^t + Bte^t + Ct^2e^t + t^3e^t \)
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You CAN Ace Differential Equations
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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