## 17Calculus Differential Equations - Exam 2

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

11

Total Points

100

Tools

Calculator

no

Formula Sheet(s)

no

Other Tools

none

Instructions:
- This exam is in four main parts, labeled sections 1-4, 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

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

$$y'' - 2y' - 3y = 0$$

$$y(t) = c_1 e^{-t} + c_2 e^{3t}$$

Problem Statement

$$y'' - 2y' - 3y = 0$$

Solution

 $$r^2 - 2r - 3 = 0$$ $$(r-3)(r+1) = 0$$ $$r=3, ~~~ r = -1$$

Solutions to the characteristic equation are real and distinct.
$$y(t) = c_1 e^{-t} + c_2 e^{3t}$$

$$y(t) = c_1 e^{-t} + c_2 e^{3t}$$

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$$y'' - 2y' + y = 0$$

Problem Statement

$$y'' - 2y' + y = 0$$

$$y(t) = c_1 e^t + c_2 t e^t$$

Problem Statement

$$y'' - 2y' + y = 0$$

Solution

 $$r^2 - 2r + 1 = 0$$ $$(r-1)^2 = 0$$ $$r = 1, r = 1$$

Solutions to the characteristic equation are real and equal.

$$y(t) = c_1 e^t + c_2 t e^t$$

$$y(t) = c_1 e^t + c_2 t e^t$$

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$$y''-2y'+5y=0$$

Problem Statement

$$y''-2y'+5y=0$$

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

Problem Statement

$$y''-2y'+5y=0$$

Solution

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

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

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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

$$y'' + 4y = e^{2x}$$

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

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 non-homogeneous 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. $$y = y_h + y_p$$.

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

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$$y''-4y = e^{2x}$$

Problem Statement

$$y''-4y = e^{2x}$$

$$y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x}$$

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 non-homogeneous (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, $$y = y_h + y_p$$

$$y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x}$$

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

Problem Statement

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

$$\displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }$$

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 $$r^2 - 2r + 1 = 0$$ $$(r-1)^2 = 0$$ $$r = 1$$ multiplicity 2 homogeneous solution $$y_h = c_1 e^x + c_2 x e^x$$

So the form of the non-homogeneous (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 non-homgeneous solutions, $$y = y_h + y_p$$.

$$\displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }$$

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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 $$y'' - 4y' + 4y = \sec^2(x)e^{2x}$$

$$y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x}$$

Problem Statement

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

Solution

Homogeneous solution first. $$r^2 - 4r + 4 = (r-2)^2 = 0 ~~~ \to ~~~ r = +2$$ 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 non-homogeneous 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.

$$y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x}$$

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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 $$y_1 = x^3$$

$$y_2 = x^{-2}$$ 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 $$u = v'$$.
$$\begin{array}{rcl} x^5 u' + 6x^4u & = & 0 \\ x^5 u' & = & -6x^4 u \\ \displaystyle{\frac{u'}{u}} & = & \displaystyle{\frac{-6}{x}} \\ \displaystyle{\frac{du}{u}} & = & \displaystyle{\frac{-6}{x}dx} \\ \ln|u| & = & -6\ln|x| + c_1 \\ u & = & c_1 x^{-6} \\ v' & = & c_1 x^{-6} \\ dv & = & c_1 x^{-6}dx \\ \displaystyle{v = \frac{c_1 x^{-5}}{-5} + k} \end{array}$$

$$\displaystyle{ y_2 = v x^3 = \left[ \frac{c_1 x^{-5}}{-5} + k \right] x^3 = \frac{c_1x^{-2}}{-5} + kx^3 }$$

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.

$$y_2 = x^{-2}$$ 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

$$y'' - 4y = 0; y(0)=0, y'(0)=1$$

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

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 $$c_2$$.
$$y(0) = c_1 + c_2 = 0 ~~~ \to ~~~ c_1 = -c_2$$
$$\begin{array}{rcl} y'(t) & = & -2c_1 e^{-2t} + 2 c_2 e^{2t} \\ y'(0) & = & -2c_1 + 2c_2 = 1 \\ & & -2(-c_2) + 2c_2 = 1 \\ & & 4c_2 = 1 \\ & & c_2 = 1/4 \\ & & c_1 = -c_2 = -1/4 \end{array}$$

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

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

Problem Statement

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

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

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
$$y(0) = c_1 + 1/8 = 1 ~~~ \to ~~~ c_1 = 1-1/8 = 7/8$$

$$\displaystyle{ y'(x) = -2c_1 \sin(2x) + 2c_2 \cos(2x) + \frac{1}{8}(2)e^{2x} }$$

$$y'(0) = 2c_2 + 1/4 = 3 ~~~ \to ~~~ 2c_2 = 3-1/4 = 11/4 ~~~ \to ~~~ c_2 = 11/8$$

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

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

Problem Statement

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

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

$$\begin{array}{rcl} r^2+r+1.25 & = & 0 \\ r^2+r+1/4 & = & -5/4+1/4 \\ (r+1/2)^2 & = & -1 \\ r+1/2 & = & \pm i \\ r & = & -1/2 \pm i \end{array}$$
$$y_h=e^{-x/2}[c_1\cos(x)+c_2\sin(x)]$$
$$\begin{array}{rcl} y_p & = & A\cos(x)+B\sin(x) \\ y'_p & = & -A\sin(x)+B\cos(x) \\ y''_p & = & -A\cos(x)-B\sin(x) \end{array}$$

 $$\displaystyle{y''_p+y'_p+\frac{5}{4}y}$$ $$\displaystyle{[-A\cos(x)-B\sin(x)]+[-A\sin(x)+B\cos(x)]+\frac{5}{4}[A\cos(x)+B\sin(x)]}$$ $$[-A+B+5A/4]\cos(x)+[-B-A+5B/4]\sin(x)=5\sin(x)$$

Equating coefficients, we have two equations and two unknowns.

cosine term sine term $$B+A/4=0$$ $$-A+B/4=5$$ $$4B+A=0$$ $$-4A+B=20$$ $$B=20+4A$$ $$4(20+4A)+A=0$$ $$80+16A+A=0$$ $$17A=-80$$ $$A=-80/17$$ $$B=20+4(-80/17)$$ $$B=(340-320)/17$$ $$B=20/17$$

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 $$c_2$$.

 $$y(0)=0$$

$$y(0)=1(c_1+0)-(80/17)(1)+0=0 ~~~ \to ~~~ c_1=80/17$$

 $$y'(0)=0$$

$$y'=e^{-x/2}[-c_1\sin(x)+c_2\cos(x)]+(80/17)\sin(x)+(20/17)\cos(x)$$
$$y'(0)=1(0+c_2)+0+20/17=0 ~~~ \to ~~~ c_2=-20/17$$

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

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You CAN Ace Differential Equations

 trig derivatives integration by parts separation of variables first-order linear homogeneous Wronskian undetermined coefficients variation of parameters

### 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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