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Differential Equations  Exam A1 
This page contains a complete differential equations exam with worked out solutions to all problems. 
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 answers.
Classify each of the following differential equations by order and type. Each question in this section is worth 3 points.
Question 1 
\(\displaystyle{ y \frac{d^2y}{dx^2} + 4\frac{dy}{dx} = \sin(x) }\)


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.
Question 2 
\(\displaystyle{ \frac{d^3 y}{dy^{3}}+4\sin(x)~y = 0 }\)


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
Solve the following initial value problems. Each question in this section is worth 5 points.
Question 3 
\( y'+2y=0; ~~~ y(0)=1 \)



\(\displaystyle{ y = e^{2x} }\)

\(\displaystyle{
\begin{array}{rcl}
y' + 2y & = & 0 \\
\frac{dy}{dx} & = & 2y \\
\frac{dy}{y} & = & 2dx \\
\int{\frac{dy}{y}} & = & \int{2dx} \\
\ln\abs{y} & = & 2x + C \\
y & = & Ae^{2x}
\end{array}
}\)
Using the initial conditions \( y(0)=1 \),
\(\displaystyle{ 1 = Ae^0 ~~~ \to ~~~ A = 1 }\)
Question 3 Final Answer 
\(\displaystyle{ y = e^{2x} }\)

Question 4 
\( y'+ty=t;~~~y(0)=2\)



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

\(\displaystyle{
\begin{array}{rcl}
y'+ty & = & t \\
& = & tty \\
& = & t(1y) \\
\frac{dy}{1y} & = & t~dt \\
\int{ \frac{dy}{1y} } & = & \int{t~dt} \\
\ln\abs{1y} & = & \frac{t^2}{2} + c \\
\frac{1}{1y} & = & Ae^{t^2/2} \\
1y & = & Be^{t^2/2} \\
y & = & 1Be^{t^2/2}
\end{array}
}\)
Using the initial condition \( y(0)=2 \),
\( 2 = 1B ~~~ \to ~~~ B = 1 \)
Question 4 Final Answer 
\(\displaystyle{ y = 1+e^{t^2/2} }\)

Find the general solution to each of the following. Each question in this section is worth 7 points.
Question 5 
\( y'=(x+1)y \)



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

\(\displaystyle{
\begin{array}{rcl}
y' & = & (x+1)y \\
\frac{dy}{y} & = & (x+1)~dx \\
\int{\frac{dy}{y}} & = & \int{(x+1)~dx} \\
\ln\abs{y} & = & \frac{x^2}{2} + x + C \\
y & = & A~\exp\left( \frac{x^2}{2}+x \right)
\end{array}
}\)
Question 5 Final Answer 
\(\displaystyle{ y = A~\exp\left( \frac{x^2}{2}+x \right) }\)

Question 6 
\( y'=yy^2 \)



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

\(\displaystyle{y' = yy^2 ~~~ \to ~~~ \frac{dy}{y(1y)} = dx }\)
Use partial fraction expansion on the left side.
\(\displaystyle{
\begin{array}{rcl}
\frac{dy}{y(1y)} & = & dx \\
\int{\frac{1}{y} + \frac{1}{1y}dy} & = & \int{1~dx} \\
\ln\abs{y}  \ln\abs{1y} & = & x + C \\
\ln \abs{\frac{y}{1y}} & = & \\
\frac{y}{1y} & = & Ae^x
\end{array}
}\)
Now we need to solve for \(y\).
\(\displaystyle{
\begin{array}{rcl}
\frac{y}{1y} & = & Ae^x \\
y & = & Ae^x (1y) \\
& = & Ae^x  yAe^x \\
y + yAe^x & = & Ae^x \\
y(1+Ae^x) & = & Ae^x \\
y & = & \frac{Ae^x}{1+Ae^x}
\end{array}
}\)
Question 6 Final Answer 
\(\displaystyle{ y = \frac{Ae^x}{1+Ae^x} }\)

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.
Question 7 
\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)



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

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 \( \psi = x^2y^2 + x^4 + y^4 + C \)
Question 7 Final Answer 
\( y = x^2y^2 + x^4 + y^4 + C \)

Question 8 
\( y+2e^x + (1+e^{x})y' = 0 \)



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

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.
\(\displaystyle{
\begin{array}{rcl}
\frac{d\mu}{dx} & = & \frac{M_y  N_x}{N}\mu \\
& = & \frac{1+e^{x}}{1+e^{x}}\mu \\
& = & \mu \\
\frac{d\mu}{\mu} & = & dx \\
ln\mu & = & x + k ~~ \text{(let k=0)}\\
\mu & = & e^x
\end{array}
}\)
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 \( \psi = ye^x +y + e^{2x} + C \)
Question 8 Final Answer 
\( y = ye^x +y + e^{2x} + C \)

Solve the following problems.
Question 9 
[5 points] Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1y^2 \).




To determine if a function satisfies a differential equation, we just plug it into the original differential equation.
\( y = \tanh(x+a) \) \( \to \) \( y' = \sech^2(x+a) \)
Let's rearrange the differential equation \( y'=1y^2 \) \( \to \) \( y' + y^2 1 = 0 \)
\(\displaystyle{
\begin{array}{rcl}
y' + y^2  1 & = & \sech^2(x+a) + \tanh^2(x+a)  1 \\
& = & \frac{1}{\cosh^2(x+a)} + \frac{\sinh^2(x+a)}{\cosh^2(x+a)}  \frac{\cosh^2(x+a)}{\cosh^2(x+a)} \\
& = & \frac{1+\sinh^2(x+a)\cosh^2(x+a)}{\cosh^2(x+a)} \\
& = & \frac{11}{\cosh^2(x+a)} = 0 ~~~ \text{[qed]}
\end{array}
}\)
Note   We used the hyperbolic identity \( \cosh^2(x+a)  \sinh^2(x+a) = 1 \)
Question 10 
[10 points] Determine the stability of each critical point for the differential equation \( y'=y^2(y^21) \).


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 to the right shows the direction field (in black) and several solutions (in color).
Question 11 
[20 points] 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 at any time. ( \(g = 32~ft/sec^2; \mu = 0.0008~ft^{1}\) )



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

The complete solution can be found on the differential equation projectile motion page with 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
\( \omega = \sqrt{\mu g} = \sqrt{32(0.0008)} = 4/25 = 0.16 \)
\( V_T = \sqrt{g/\mu} = \sqrt{32/0.0008} = 200 \)
\( v(t) = 200\tanh(0.16t) \)
Question 11 Final Answer 
\( v(t) = 200\tanh(0.16t)\)

Question 12 
[ 15 points ] 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?



\(\displaystyle{ \frac{dQ}{dt} = 1000  \frac{5Q}{20+5t} }\);
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.

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 \(\displaystyle{ \frac{dQ}{dt} = rate~in  rate~out }\).
\(\displaystyle{ rate~in = \frac{100g}{gal} \cdot \frac{10 gal}{min} = 1000 g/min }\)
\(\displaystyle{ rate~out = \frac{Q~g}{V~gal} \cdot \frac{5gal}{min} = \frac{5Q}{20+5t} }\)
Question 12 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.
