This page contains a complete differential equations exam with worked out solutions to all problems.
Note  This is not exam A1. It 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  12 
Total Points  100 
Tools  

Calculator  no 
Formula Sheet(s)  no 
Other Tools  none 
Instructions:
 This exam is in five main parts, labeled sections 15, 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
Classify each of the following differential equations by order and type. Each question in this section is worth 3 points.
\(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\)
Problem Statement 

Classify the differential equation \(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\) by order and type.
Final Answer 

second order, nonlinear, nonhomogeneous
Problem Statement 

Classify the differential equation \(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\) by order and type.
Solution 

The differential equation is
 second order due to the \(\displaystyle{ \frac{d^2y}{dx^2} }\) term
 nonlinear due to multiplication of y in \(\displaystyle{ \left[ y\frac{d^2y}{dx^2} \right] }\)
 nonhomogeneous due to the \( \sin(x) \) term.
Final Answer 

second order, nonlinear, nonhomogeneous
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\(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\)
Problem Statement 

Classify the differential equation \(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\) by order and type.
Final Answer 

third order, linear, homogeneous
Problem Statement 

Classify the differential equation \(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\) by order and type.
Solution 

The differential equation is
 third order due to the \(\displaystyle{ \frac{d^3 y}{dy^3} }\) term
 linear
 homogeneous since the right side is equal to zero
Final Answer 

third order, linear, homogeneous
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Section 2
Solve the following initial value problems. Each question in this section is worth 5 points.
\( y'+2y=0; ~~~ y(0)=1 \)
Problem Statement 

Solve the initial value problem \( y'+2y=0; ~~~ y(0)=1 \)
Final Answer 

\( y = e^{2x} \)
Problem Statement 

Solve the initial value problem \( y'+2y=0; ~~~ y(0)=1 \)
Solution 

\(\begin{array}{rcl} y' + 2y & = & 0 \\ \displaystyle{\frac{dy}{dx}} & = & 2y \\ \displaystyle{\frac{dy}{y}} & = & 2dx \\ \displaystyle{\int{\frac{dy}{y}}} & = & \int{2dx} \\ \ln\abs{y} & = & 2x + C \\ y & = & Ae^{2x} \end{array}\)
Using the initial conditions \( y(0)=1 \),
\( 1 = Ae^0 ~~~ \to ~~~ A = 1 \)
Final Answer 

\( y = e^{2x} \)
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\( y'+ty=t; ~~~ y(0)=2\)
Problem Statement 

Solve the initial value problem \( y'+ty=t; ~~~ y(0)=2\)
Final Answer 

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

Solve the initial value problem \( y'+ty=t; ~~~ y(0)=2\)
Solution 

\(\begin{array}{rcl} y'+ty & = & t \\ & = & tty \\ & = & t(1y) \\ \displaystyle{\frac{dy}{1y}} & = & t~dt \\ \displaystyle{\int{ \frac{dy}{1y} }} & = & \int{t~dt} \\ \ln\abs{1y} & = & \displaystyle{\frac{t^2}{2} + c} \\ \displaystyle{\frac{1}{1y}} & = & Ae^{t^2/2} \\ 1y & = & Be^{t^2/2} \\ y & = & 1Be^{t^2/2} \end{array}\)
Using the initial condition \( y(0)=2 \),
\( 2 = 1B ~~~ \to ~~~ B = 1 \)
Final Answer 

\(\displaystyle{ y = 1 + e^{t^2/2} }\)
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Section 3
Find the general solution to each of the following. Each question in this section is worth 7 points.
\( y' = (x+1)y \)
Problem Statement 

Find the general solution to \( y' = (x+1)y \)
Final Answer 

\(\displaystyle{ y = A~\exp\left( \frac{x^2}{2} + x \right) }\)
Problem Statement 

Find the general solution to \( y' = (x+1)y \)
Solution 

\(\begin{array}{rcl} y' & = & (x+1)y \\ \displaystyle{\frac{dy}{y}} & = & (x+1)~dx \\ \displaystyle{\int{\frac{dy}{y}}} & = & \int{(x+1)~dx} \\ \ln\abs{y} & = & \displaystyle{\frac{x^2}{2} + x + C} \\ y & = & \displaystyle{A~\exp\left( \frac{x^2}{2}+x \right)} \end{array}\)
Final Answer 

\(\displaystyle{ y = A~\exp\left( \frac{x^2}{2} + x \right) }\)
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\( y' = y  y^2 \)
Problem Statement 

Find the general solution to \( y' = y  y^2 \)
Final Answer 

\(\displaystyle{ y = \frac{Ae^x}{1+Ae^x} }\)
Problem Statement 

Find the general solution to \( y' = y  y^2 \)
Solution 

\(\displaystyle{y' = yy^2 ~~~ \to ~~~ \frac{dy}{y(1y)} = dx }\)
Use partial fraction expansion on the left side.
\(\begin{array}{rcl} \displaystyle{\frac{dy}{y(1y)}} & = & dx \\ \displaystyle{\int{\frac{1}{y} + \frac{1}{1y}dy}} & = & \int{1~dx} \\ \ln\abs{y}  \ln\abs{1y} & = & x + C \\ \displaystyle{\ln \abs{\frac{y}{1y}}} & = & \\ \displaystyle{\frac{y}{1y}} & = & Ae^x \end{array}\)
Now we need to solve for \(y\).
\(\begin{array}{rcl} \displaystyle{\frac{y}{1y}} & = & Ae^x \\ y & = & Ae^x (1y) \\ & = & Ae^x  yAe^x \\ y + yAe^x & = & Ae^x \\ y(1+Ae^x) & = & Ae^x \\ y & = & \displaystyle{\frac{Ae^x}{1+Ae^x}} \end{array}\)
Final Answer 

\(\displaystyle{ y = \frac{Ae^x}{1+Ae^x} }\)
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Section 4
Determine if the following equations are exact. If not, calculate the integrating factor to make them exact and then solve each equation. Each problem in this section is worth 10 points.
\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)
Problem Statement 

\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)
Final Answer 

\( y = x^2y^2 + x^4 + y^4 + C \)
Problem Statement 

\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)
Solution 

First, check for exactness. Rewriting the equation in differential form, we have \( (2xy^2 + 4x^3)dx + (2x^2y +4y^3)dy = 0 \)
\( M(x,y) = 2xy^2 + 4x^3 \) \( \to \) \( M_y = (2x)(2y) + 0 = 4xy \)
\( N(x,y) = 2x^2y+4y^3 \) \( \to \) \( N_x = 2y(2x)+ 0 = 4xy \)
Since \( M_y = N_x \), the differential equation is exact and so we can solve it.
\( \psi_x = M(x,y) = 2xy^2+4x^3 \) \( \to \) \( \psi = 2y^2 (x^2/2) + (4x^4)/4 + h(y) = \) \( x^2y^2 + x^4 + h(y) \)
\( \psi_y = N(x,y) = 2x^2y+4y^3 \) \( \to \) \( \psi = 2x^2 (y^2/2) + (4y^4)/4 + g(x) = \) \( x^2y^2 + y^4 + g(x) \)
Equating these results, we get \( \psi = x^2y^2 + x^4 + y^4 + C \)
Final Answer 

\( y = x^2y^2 + x^4 + y^4 + C \)
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\( y+2e^x + (1+e^{x})y' = 0 \)
Problem Statement 

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

\( y = ye^x +y + e^{2x} + C \)
Problem Statement 

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

First we check for exactness. Rewriting the differential equation, we have \( (y+2e^x)dx + (1+e^{x})dy = 0 \).
\( M(x,y) = y+2e^x \) \( \to \) \( M_y = 1 \)
\( N(x,y) = 1+e^{x} \) \( \to \) \( N_x = e^{x} \)
Since \( M_y \neq N_x \), this equation is not exact. We need to find an integrating factor.
\(\begin{array}{rcl} \displaystyle{\frac{d\mu}{dx}} & = & \displaystyle{\frac{M_y  N_x}{N}\mu} \\ & = & \displaystyle{\frac{1+e^{x}}{1+e^{x}}\mu} \\ & = & \mu \\ \displaystyle{\frac{d\mu}{\mu}} & = & dx \\ ln\mu & = & x + k ~~ \text{(let k=0)}\\ \mu & = & e^x \end{array}\)
So our integrating factor is \( \mu=e^x \). Multiplying the differential equation by this factor, we get
\( e^x[t+2e^x] + e^x[1+e^{x}]y' = 0 \) \( \to \) \( (ye^x+2e^{2x})dx + (e^x+1)dy = 0 \)
Let's check that we now have an exact equation.
\( M(x,y) = ye^x + 2e^{2x} \) \( \to \) \( M_y = e^x \)
\( N(x,y) = e^x+1 \) \( \to \) \( N_x = e^x \)
Since we now have an exact equation, we can solve it.
\( \psi_x = M(x,y) = ye^x + 2e^{2x} \) \( \to \) \( \psi = ye^x + e^{2x} + h(y) \)
\( \psi_y = N(x,y) = e^x+1 \) \( \to \) \( \psi = ye^x + y + g(x) \)
Equating we get \( \psi = ye^x +y + e^{2x} + C \)
Final Answer 

\( y = ye^x +y + e^{2x} + C \)
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Section 5
Solve the following problems.
Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1y^2 \).
Problem Statement 

Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1y^2 \).
Hint 

To determine if a function satisfies a differential equation, plug it into the original differential equation and check that the equation holds.
Problem Statement 

Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1y^2 \).
Hint 

To determine if a function satisfies a differential equation, plug it into the original differential equation and check that the equation holds.
Solution 

\( y = \tanh(x+a) \) \( \to \) \( y' = \sech^2(x+a) \)
Let's rearrange the differential equation
\( y'=1y^2 \) \( \to \) \( y' + y^2 1 = 0 \)
\( y' + y^2  1 \) 
\( \sech^2(x+a) + \tanh^2(x+a)  1 \) 
\( \displaystyle{\frac{1}{\cosh^2(x+a)} + \frac{\sinh^2(x+a)}{\cosh^2(x+a)}  \frac{\cosh^2(x+a)}{\cosh^2(x+a)}} \) 
\( \displaystyle{\frac{1+\sinh^2(x+a)\cosh^2(x+a)}{\cosh^2(x+a)}} \) 
Use the hyperbolic identity \( \cosh^2(x+a)  \sinh^2(x+a) = 1 \) 
\( \displaystyle{\frac{11}{\cosh^2(x+a)} = 0} ~~~ \text{[qed]} \) 
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Determine the stability of each critical point for the differential equation \( y'=y^2(y^21) \).
Problem Statement 

Determine the stability of each critical point for the differential equation \( y'=y^2(y^21) \).
Solution 

This equation has 3 critical points, \( y=0, y=1, y=1 \), since the derivative is zero at those points. For \( y>1\) and \( y<1\), the slope goes off to infinity and is therefore unstable. For \( 1 < y < 1 \), the system is stable with asymptotes at the critical points.
The plot shows the direction field (in black) and several solutions (in color).
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A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant \(\mu\)), determine his velocity as a function of time. \(\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{1} \right]\)
Problem Statement 

A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant \(\mu\)), determine his velocity as a function of time. \(\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{1} \right]\)
Final Answer 

\(v(t) = 200\tanh(0.16t)\)
Problem Statement 

A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant \(\mu\)), determine his velocity as a function of time. \(\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{1} \right]\)
Solution 

We use the result \( v(t) = V_T \tanh(\omega t ) \) where \( \omega = \sqrt{\mu g} \) and \( V_T = \sqrt{g/\mu} \).
Plugging in the given numbers we have
\( \omega = \sqrt{\mu g} = \sqrt{32(0.0008)} = 4/25 = 0.16 \)
\( V_T = \sqrt{g/\mu} = \sqrt{32/0.0008} = 200 \)
\( v(t) = 200\tanh(0.16t) \)
Final Answer 

\(v(t) = 200\tanh(0.16t)\)
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A 100 gallon tank is being filled with a solution containing 100 grams per gallon of chlorine. The tank is being filled at the rate of 10 gallons per minute. Initially the tank contains 20 gallons with chlorine content of 200 grams. The wellmixed solution leaves the tank at a rate of 5 gallons per minute. Set up but do not solve the initial value problem that models this. If they fail to stop the flow in when the tank is full, what will the eventual chlorine content be?
Problem Statement 

A 100 gallon tank is being filled with a solution containing 100 grams per gallon of chlorine. The tank is being filled at the rate of 10 gallons per minute. Initially the tank contains 20 gallons with chlorine content of 200 grams. The wellmixed solution leaves the tank at a rate of 5 gallons per minute. Set up but do not solve the initial value problem that models this. If they fail to stop the flow in when the tank is full, what will the eventual chlorine content be?
Final Answer 

\(\displaystyle{ \frac{dQ}{dt} = 1000  \frac{5Q}{20+5t} }\); As \( t \to \infty \) the concentration in the tank will be the same as is pouring in, i.e \( 100g/gal \) or \( 10000g \) in the entire tank.
Problem Statement 

A 100 gallon tank is being filled with a solution containing 100 grams per gallon of chlorine. The tank is being filled at the rate of 10 gallons per minute. Initially the tank contains 20 gallons with chlorine content of 200 grams. The wellmixed solution leaves the tank at a rate of 5 gallons per minute. Set up but do not solve the initial value problem that models this. If they fail to stop the flow in when the tank is full, what will the eventual chlorine content be?
Solution 

Flowing In: \( 100g/gal \) flowing in at a rate of \( 10gal/min\).
Flowing Out: Concentation of \( Q~g/V~gal \) flowing out at a rate of \( 5 gal/min \).
Initial State: \( 20 gal\) with \( 200g\) of chlorine.
Volume In Tank: initial amt + (flow in  flow out)(time elapsed), so \(V= 20 gal + (10gal/min  5/min)t = 20+5t\)
We need to set up the equation \(\displaystyle{ \frac{dQ}{dt} = rate~in  rate~out }\).
\(\displaystyle{ rate~in = \frac{100g}{gal} \cdot \frac{10 gal}{min} = 1000 g/min }\)
\(\displaystyle{ rate~out = \frac{Q~g}{V~gal} \cdot \frac{5gal}{min} = \frac{5Q}{20+5t} }\)
Final Answer 

\(\displaystyle{ \frac{dQ}{dt} = 1000  \frac{5Q}{20+5t} }\); As \( t \to \infty \) the concentration in the tank will be the same as is pouring in, i.e \( 100g/gal \) or \( 10000g \) in the entire tank.
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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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