## 17Calculus Differential Equations - Exam 1

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 1-5, with different instructions for each section.
- 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).

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.

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.

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.

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

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 $$y'+2y=0; ~~~ y(0)=1$$

$$y = e^{-2x}$$

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 $$y(0)=1$$,
$$1 = Ae^0 ~~~ \to ~~~ A = 1$$

$$y = e^{-2x}$$

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$$y'+ty=t; ~~~ y(0)=2$$

Problem Statement

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

$$\displaystyle{ y = 1 + e^{-t^2/2} }$$

Problem Statement

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

Solution

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

$$\displaystyle{ y = 1 + e^{-t^2/2} }$$

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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 $$y' = (x+1)y$$

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

Problem Statement

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

Solution

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

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

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$$y' = y - y^2$$

Problem Statement

Find the general solution to $$y' = y - y^2$$

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

Problem Statement

Find the general solution to $$y' = y - y^2$$

Solution

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

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

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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

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

$$y = x^2y^2 + x^4 + y^4 + C$$

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 $$\psi = x^2y^2 + x^4 + y^4 + C$$

$$y = x^2y^2 + x^4 + y^4 + C$$

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$$y+2e^x + (1+e^{-x})y' = 0$$

Problem Statement

$$y+2e^x + (1+e^{-x})y' = 0$$

$$y = ye^x +y + e^{2x} + C$$

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 $$\psi = ye^x +y + e^{2x} + C$$

$$y = ye^x +y + e^{2x} + C$$

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Section 5

Solve the following problems.

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

Problem Statement

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

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 $$y'=1-y^2$$.

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
$$y'=1-y^2$$ $$\to$$ $$y' + y^2 -1 = 0$$

 $$y' + y^2 - 1$$ $$\sech^2(x+a) + \tanh^2(x+a) - 1$$ $$\displaystyle{\frac{1}{\cosh^2(x+a)} + \frac{\sinh^2(x+a)}{\cosh^2(x+a)} - \frac{\cosh^2(x+a)}{\cosh^2(x+a)}}$$ $$\displaystyle{\frac{1+\sinh^2(x+a)-\cosh^2(x+a)}{\cosh^2(x+a)}}$$ Use the hyperbolic identity $$\cosh^2(x+a) - \sinh^2(x+a) = 1$$ $$\displaystyle{\frac{1-1}{\cosh^2(x+a)} = 0} ~~~ \text{[qed]}$$

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Determine the stability of each critical point for the differential equation $$y'=y^2(y^2-1)$$.

Problem Statement

Determine the stability of each critical point for the differential equation $$y'=y^2(y^2-1)$$.

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. $$\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{-1} \right]$$

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

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
$$\omega = \sqrt{\mu g} = \sqrt{32(0.0008)} = 4/25 = 0.16$$
$$V_T = \sqrt{g/\mu} = \sqrt{32/0.0008} = 200$$
$$v(t) = -200\tanh(0.16t)$$

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

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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 well-mixed 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 well-mixed 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.

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 well-mixed 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 $$\displaystyle{ \frac{dQ}{dt} = rate~in - rate~out }$$.

$$\displaystyle{ rate~in = \frac{100g}{gal} \cdot \frac{10 gal}{min} = 1000 g/min }$$

$$\displaystyle{ rate~out = \frac{Q~g}{V~gal} \cdot \frac{5gal}{min} = \frac{5Q}{20+5t} }$$

$$\displaystyle{ \frac{dQ}{dt} = 1000 - \frac{5Q}{20+5t} }$$; 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

 hyperbolic derivatives integration using partial fractions separation of variables exact equations projectile motion chemical concentration in fluids

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

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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