This page contains a complete differential equations exam with worked out solutions to all problems.
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IMPORTANT 
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Exam Details  

Time  1 hour 
Questions  11 
Total Points  100 
Tools  

Calculator  no 
Formula Sheet(s)  no 
Other Tools  none 
Instructions:
 This exam is in four main parts, labeled sections 14, 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 of each of the following differential equations. Each question in this section is worth 5 points.
\( y''  2y'  3y = 0 \)
Problem Statement 

\( y''  2y'  3y = 0 \)
Final Answer 

\( y(t) = c_1 e^{t} + c_2 e^{3t} \)
Problem Statement 

\( y''  2y'  3y = 0 \)
Solution 

\( r^2  2r  3 = 0 \) 
\( (r3)(r+1) = 0 \) 
\( r=3, ~~~ r = 1 \) 
Solutions to the characteristic equation are real and distinct.
\( y(t) = c_1 e^{t} + c_2 e^{3t} \)
Final Answer 

\( y(t) = c_1 e^{t} + c_2 e^{3t} \) 
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\( y''  2y' + y = 0 \)
Problem Statement 

\( y''  2y' + y = 0 \)
Final Answer 

\( y(t) = c_1 e^t + c_2 t e^t \)
Problem Statement 

\( y''  2y' + y = 0 \)
Solution 

\( r^2  2r + 1 = 0 \) 
\( (r1)^2 = 0 \) 
\( r = 1, r = 1 \) 
Solutions to the characteristic equation are real and equal.
\( y(t) = c_1 e^t + c_2 t e^t \)
Final Answer 

\( y(t) = c_1 e^t + c_2 t e^t \) 
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\( y''2y'+5y=0 \)
Problem Statement 

\( y''2y'+5y=0 \)
Final Answer 

\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \)
Problem Statement 

\( y''2y'+5y=0 \)
Solution 

\(\begin{array}{rcl} r^2  2r + 5 & = & 0 \\ r^2  2r + 1 & = & 5 + 1 \\ (r1)^2 & = & 4 \\ r1 & = & \pm 2i \\ r & = & 1 \pm 2i \end{array}\)
Solutions to the characteristic equation are complex.
\( y(t) = e^t [c_1 \cos(2t) + c_2 \sin(2t) ] \)
Final Answer 

\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \) 
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Section 2
Find the general solution to each of the following differential equations using the method of undetermined coefficients. Each question in this section is worth 10 points.
\( y'' + 4y = e^{2x} \)
Problem Statement 

\( y'' + 4y = e^{2x} \)
Final Answer 

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \)
Problem Statement 

\( y'' + 4y = e^{2x} \)
Solution 

First, we need to solve the homogeneous equation \( y'' + 4y = 0 \).
\( r^2+4 = 0 ~~~ \to ~~~ r^2 = 4 ~~~ \to ~~~ r = \pm 2i \)
\( y_h = c_1 \cos(2x) + c_2 \sin(2x) \)
So the form of the nonhomogeneous solution is \( y_p(x) = Ae^{2x} \)
\( y'_p(x) = 2Ae^{2x} ~~~~~~ y''_p(x) = 4Ae^{2x} \)
\(\begin{array}{rcl} y''_p + 4y_p & = & e^{2x} \\ 4Ae^{2x} + 4(Ae^{2x}) & = & e^{2x} \\ 8Ae^{2x} & = & e^{2x} \\ 8A & = & 1 \\ A & = & 1/8 \end{array}\)
Combining the two solutions gives us the final answer, i.e. \( y = y_h + y_p \).
Final Answer 

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \) 
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\( y''4y = e^{2x} \)
Problem Statement 

\( y''4y = e^{2x} \)
Final Answer 

\( y = c_1e^{2x} + c_2 e^{2x} + (x/4)e^{2x} \)
Problem Statement 

\( y''4y = e^{2x} \)
Solution 

First, we need to find the solution to the homogeneous equation, \( y''  4y = 0 \).
\( y''  4y = 0 \) 
characteristic equation 
\( r^2  4 = 0 \) 
\( r^2 = 4 \) 
\( r = \pm 2 \) 
homogeneous solution 
\( y_h(x) = c_1 e^{2x} + c_2 e^{2x} \) 
Now the form of the nonhomogeneous (particular) solution is \( y_p(x) = Axe^{2x} \).
\( y_p(x) = Axe^{2x} \) 
\( y'_p = Ax(2e^{2x}) + Ae^{2x} = (2Ax+A)e^{2x} \) 
\( y''_p = (2Ax+A)(2e^{2x}) + e^{2x}(2A) = (4Ax+2A+2A)e^{2x} = (4Ax+4A)e^{2x} \) 
plug these expressions into the original differential equation 
\( y''_p  4y_p = e^{2x} \) 
\( (4Ax+4A)e^{2x}  4(Axe^{2x}) = e^{2x} \) 
\( 4Ae^{2x} = e^{2x} \) 
\( 4A = 1 \) 
\( A = 1/4 \) 
particular solution 
\( (x/4)e^{2x} \) 
Combining the homogeneous and particular solutions gives us the complete solution, \( y = y_h + y_p \)
Final Answer 

\( y = c_1e^{2x} + c_2 e^{2x} + (x/4)e^{2x} \) 
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\( y''  2y' + y = (2x+1)e^x \)
Problem Statement 

\( y''  2y' + y = (2x+1)e^x \)
Final Answer 

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\)
Problem Statement 

\( y''  2y' + y = (2x+1)e^x \)
Solution 

As before, we need to find the solutions to the homogeneous equation.
\( y''2y'+y = 0 \) 
characteristic equation 
\( r^2  2r + 1 = 0 \) 
\( (r1)^2 = 0 \) 
\( r = 1 \) multiplicity 2 
homogeneous solution 
\( y_h = c_1 e^x + c_2 x e^x \) 
So the form of the nonhomogeneous (particular) solution is \( y_p = (Ax+B)e^x (x^2) = (Ax^3+Bx^2)e^x \)
\( y_p = (Ax^3+Bx^2)e^x \) 
\( y'_p = (Ax^3+Bx^2)e^x + e^x(3Ax^2+2Bx) \) 
\( y'_p = (Ax^3 + 3Ax^2 + Bx^2 + 2Bx)e^x \) 
\( y''_p = (Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + e^x(3Ax^2+6Ax+2Bx+2B) \) 
\( y''_p = (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \) 
Now plug these expressions into the original differential equation. 
\( y''_p  2y'_p + y_p \) 
\( (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \) 
\( 2(Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + (Ax^3 + Bx^2)e^x \) 
\( (6Ax + 2B)e^x = (2x+1)e^x \) 
Equating coefficients in the last equation yields 
\( 2B = 1 ~~~ \to ~~~ B = 1/2 \) 
\( 6A = 2 ~~~ \to ~~~ A = 1/3 \) 
particular solution 
\(\displaystyle{ y_p = \frac{x^3}{3} + \frac{x^2}{2} }\) 
The final answer combines the results from the homogeneous and nonhomgeneous solutions, \( y = y_h + y_p \).
Final Answer 

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\) 
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Section 3
Solve the following problems.
Use variation of parameters to find the general solution of \( y''  4y' + 4y = \sec^2(x)e^{2x} \)
Problem Statement 

Use variation of parameters to find the general solution of \( y''  4y' + 4y = \sec^2(x)e^{2x} \)
Final Answer 

\( y = [ \ln\cos(x) + c_2 x + c_1 ] e^{2x} \)
Problem Statement 

Use variation of parameters to find the general solution of \( y''  4y' + 4y = \sec^2(x)e^{2x} \)
Solution 

Homogeneous solution first. \( r^2  4r + 4 = (r2)^2 = 0 ~~~ \to ~~~ r = +2 \) multiplicity \(2\)
\( y_h = c_1 e^{2x} + c_2 x e^{2x} \)
Now, find the Wronskian.
\(\displaystyle{W = \begin{vmatrix} y_{h1} & y_{h2} \\ y'_{h1} & y'_{h2} \end{vmatrix} = }\) \(\displaystyle{\begin{vmatrix} e^{2x} & xe^{2x} \\ 2e^{2x} & (2x+1)e^{2x} \end{vmatrix} = }\) \(\displaystyle{ (2x+1)e^{4x}  2xe^{4x} = e^{4x} }\)
Using variation of parameters, our nonhomogeneous solution looks like \( y=u_1(x)e^{2x} + u_2(x)xe^{2x} \).
Setting up the equations for \( u'_1(x) \) and \( u'_2(x) \), we have
\(\displaystyle{u'_1 = \frac{\begin{vmatrix} 0 & xe^{2x} \\ \sec^2 xe^{2x} & (2x+1)e^{2x} \end{vmatrix} }{W} = \frac{x\sec^2 x e^{4x}}{e^{4x}} = x\sec^2 x }\)
\(\displaystyle{u'_2 = \frac{\begin{vmatrix} e^{2x} & 0 \\ 2e^{2x} & \sec^2xe^{2x} \end{vmatrix} }{W} = \frac{\sec^2xe^{4x}}{e^{4x}} = \sec^2 x }\)
So we need to solve the two equations
\(\displaystyle{ u'_1 = x \sec^2 x ~~~ \to ~~~ u_1 = \int{x\sec^2 x ~ dx} }\)
\(\displaystyle{ u'_2 = \sec^2 x ~~~ \to ~~~ u_2 = \int{\sec^2 x~dx} }\)
The first integral is solved in the practice problems on the integration by parts page. So we just give the result here.
\( \displaystyle{ u_1 = \int{x\sec^2 x ~ dx} = x\tan(x)\ln\cos(x)+k_1 }\)
The second one is easy, since \( (\tan(x))' = \sec^2(x) \)
\(\displaystyle{ u_2 = \int{\sec^2 x~dx} = \tan(x)+k_2 }\)
To get our final answer, we substitute these results into the equation \( y=u_1e^{2x} + u_2xe^{2x} \) and simplify.
We also let \( c_1 = A+k_1\) and \( c_2 = B+k_2 \) to combine the constants.
Final Answer 

\( y = [ \ln\cos(x) + c_2 x + c_1 ] e^{2x} \) 
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Determine the second fundamental solution of \( x^2y''  6y = 0 \), given \( y_1 = x^3 \)
Problem Statement 

Determine the second fundamental solution of \( x^2y''  6y = 0 \), given \( y_1 = x^3 \)
Final Answer 

\( y_2 = x^{2} \) is the second fundamental solution.
Problem Statement 

Determine the second fundamental solution of \( x^2y''  6y = 0 \), given \( y_1 = x^3 \)
Solution 

We will use the technique of Reduction of Order, where \( y_2 = vy_1 \) and \(v\) is a function of \(x\).
\( y_2 = vy_1 \) 
\( y'_2 = v(3x^2) + x^3 v' \) 
\( y''_2 = v(6x) + 3x^2 v + x^3 v'' + 3x^2 v' \) 
\( y''_2 = x^3 v'' + 6x^2 v' + 6xv \) 
\( x^2 y''_2  6y_2 = x^2[ x^3 v'' + 6x^2 v' + 6xv]  6vx^3 \) 
\( x^5 v'' + 6x^4 v' = 0 \) 
Let \( u = v' \).
\(\begin{array}{rcl} x^5 u' + 6x^4u & = & 0 \\ x^5 u' & = & 6x^4 u \\ \displaystyle{\frac{u'}{u}} & = & \displaystyle{\frac{6}{x}} \\ \displaystyle{\frac{du}{u}} & = & \displaystyle{\frac{6}{x}dx} \\ \lnu & = & 6\lnx + c_1 \\ u & = & c_1 x^{6} \\ v' & = & c_1 x^{6} \\ dv & = & c_1 x^{6}dx \\ \displaystyle{v = \frac{c_1 x^{5}}{5} + k} \end{array}\)
\(\displaystyle{ y_2 = v x^3 = \left[ \frac{c_1 x^{5}}{5} + k \right] x^3 = \frac{c_1x^{2}}{5} + kx^3 }\)
let \( c = c_1 / (5) \)
\( y_2 = c/x^2 + kx^3 \)
Since \( kx^3 \) is a multiple of \( y_1 \) it is already included in \( y_1 \), so we just drop that term.
Final Answer 

\( y_2 = x^{2} \) is the second fundamental solution. 
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Section 4
Solve the following initial value problems.
\( y''  4y = 0; y(0)=0, y'(0)=1 \)
Problem Statement 

\( y''  4y = 0; y(0)=0, y'(0)=1 \)
Final Answer 

\(\displaystyle{ y(t) = \frac{1}{4} e^{2t} + \frac{1}{4} e^{2t} }\)
Problem Statement 

\( y''  4y = 0; y(0)=0, y'(0)=1 \)
Solution 

\( r^2  4 = 0 ~~~ \to ~~~ r^2 = 4 ~~~ \to ~~~ r = \pm 2 \)
\( y = c_1 e^{2t} + c_2 e^{2t} \)
Use the initial conditions to find the constants \( c_1 \) and \( c_2 \).
\( y(0) = c_1 + c_2 = 0 ~~~ \to ~~~ c_1 = c_2 \)
\(\begin{array}{rcl} y'(t) & = & 2c_1 e^{2t} + 2 c_2 e^{2t} \\ y'(0) & = & 2c_1 + 2c_2 = 1 \\ & & 2(c_2) + 2c_2 = 1 \\ & & 4c_2 = 1 \\ & & c_2 = 1/4 \\ & & c_1 = c_2 = 1/4 \end{array}\)
Final Answer 

\(\displaystyle{ y(t) = \frac{1}{4} e^{2t} + \frac{1}{4} e^{2t} }\) 
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\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)
Problem Statement 

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)
Final Answer 

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\)
Problem Statement 

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)
Solution 

From the previous problem, we have the general solution \(\displaystyle{ y = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{8} e^{2x}}\)
Using the initial conditions to find the constants \(c_1\) and \(c_2\), we have
\( y(0) = c_1 + 1/8 = 1 ~~~ \to ~~~ c_1 = 11/8 = 7/8 \)
\(\displaystyle{ y'(x) = 2c_1 \sin(2x) + 2c_2 \cos(2x) + \frac{1}{8}(2)e^{2x} }\)
\( y'(0) = 2c_2 + 1/4 = 3 ~~~ \to ~~~ 2c_2 = 31/4 = 11/4 ~~~ \to ~~~ c_2 = 11/8 \)
Final Answer 

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\) 
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\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)
Problem Statement 

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)
Final Answer 

\(\displaystyle{y(x)=e^{x/2}\left[\frac{80}{17}\cos(x)\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\)
steady state occurs as \(x\to\infty\); so \(e^{x/2}\to 0\) and \(\displaystyle{y_{ss}(x)=\frac{80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)
Problem Statement 

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)
Solution 

\(\begin{array}{rcl} r^2+r+1.25 & = & 0 \\ r^2+r+1/4 & = & 5/4+1/4 \\ (r+1/2)^2 & = & 1 \\ r+1/2 & = & \pm i \\ r & = & 1/2 \pm i \end{array}\)
\(y_h=e^{x/2}[c_1\cos(x)+c_2\sin(x)]\)
\(\begin{array}{rcl} y_p & = & A\cos(x)+B\sin(x) \\ y'_p & = & A\sin(x)+B\cos(x) \\ y''_p & = & A\cos(x)B\sin(x) \end{array}\)
\( \displaystyle{y''_p+y'_p+\frac{5}{4}y} \) 
\( \displaystyle{[A\cos(x)B\sin(x)]+[A\sin(x)+B\cos(x)]+\frac{5}{4}[A\cos(x)+B\sin(x)]} \) 
\( [A+B+5A/4]\cos(x)+[BA+5B/4]\sin(x)=5\sin(x) \) 
Equating coefficients, we have two equations and two unknowns.
cosine term  sine term  

\(B+A/4=0\)  \(A+B/4=5\)  
\(4B+A=0\)  \(4A+B=20\)  
\(B=20+4A\)  
\(4(20+4A)+A=0\)  
\(80+16A+A=0\)  
\(17A=80\)  
\(A=80/17 \)  
\(B=20+4(80/17)\)  
\(B=(340320)/17\)  
\(B=20/17\) 
So far, we have \(\displaystyle{y=e^{x/2}[c_1\cos(x)+c_2\sin(x)]\frac{80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)
Now we use the initial conditions to determine the constants \(c_1\) and \(c_2\).
\(y(0)=0\) 
\(y(0)=1(c_1+0)(80/17)(1)+0=0 ~~~ \to ~~~ c_1=80/17\)
\(y'(0)=0\) 
\(y'=e^{x/2}[c_1\sin(x)+c_2\cos(x)]+(80/17)\sin(x)+(20/17)\cos(x)\)
\(y'(0)=1(0+c_2)+0+20/17=0 ~~~ \to ~~~ c_2=20/17\)
Final Answer 

\(\displaystyle{y(x)=e^{x/2}\left[\frac{80}{17}\cos(x)\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\) 
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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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Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

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