17Calculus Differential Equations - Exam 3

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

7

Total Points

100

Tools

Calculator

no

Formula Sheet(s)

no

Other Tools

none

Instructions:
- This exam is in three main parts, labeled sections 1-3, 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 for each of the following differential equations. Each question in this section is worth 5 points.

$$y^{(3)}-4y' = 0$$

Problem Statement

Find the general solution to $$y^{(3)}-4y' = 0$$

$$y = A + Be^{-2t} + Ce^{2t}$$

Problem Statement

Find the general solution to $$y^{(3)}-4y' = 0$$

Solution

 $$r^3 - 4r = r(r^2-4) = 0$$ $$r=0; ~~~ r^2-4=0$$ $$r^2 = 4; ~~~ r = \pm 2$$ $$y = A + Be^{-2t} + Ce^{2t}$$

$$y = A + Be^{-2t} + Ce^{2t}$$

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$$y^{(4)} + 2y'' + y = 0$$

Problem Statement

Find the general solution to $$y^{(4)} + 2y'' + y = 0$$

$$y = (A+Bt)\cos(t) + (C+Dt)\sin(t)$$

Problem Statement

Find the general solution to $$y^{(4)} + 2y'' + y = 0$$

Solution

$$\begin{array}{rcl} r^4 + 2r^2 + 1 & = & 0 \\ (r^2+1)^2 & = & 0 \\ r^2 + 1 & = & 0 ~~~ \text{multiplicity 2} \\ r^2 & = & -1 \\ r = \pm i \end{array}$$
$$y = (A+Bt)\cos(t) + (C+Dt)\sin(t)$$

$$y = (A+Bt)\cos(t) + (C+Dt)\sin(t)$$

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$$y^{(3)} - 3y''+3y' - y = 0$$

Problem Statement

Find the general solution to $$y^{(3)} - 3y''+3y' - y = 0$$

$$y = (A+Bt+Ct^2)e^t$$

Problem Statement

Find the general solution to $$y^{(3)} - 3y''+3y' - y = 0$$

Solution

$$\begin{array}{rcl} r^3 - 3r^2 + 3r - 1 & = & 0 \\ (r-1)(r^2-2r+1) & = & 0 \\ (r-1)(r-1)^2 & = & 0 \\ (r-1)^3 & = & 0 \\ r-1 & = & 0 ~~~ \text{X}3^* \\ r & = & 1 \end{array}$$
*Multiplicity 3
$$y = (A+Bt+Ct^2)e^t$$

$$y = (A+Bt+Ct^2)e^t$$

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Section 2

Find the general solution for each of the following differential equations. Use the method of undetermined coefficients to find the particular solutions. Each question in this section is worth 15 points.

$$y^{(3)} -4y' = 4e^{2t}$$

Problem Statement

Use the method of undetermined coefficients to find the general solution to $$y^{(3)} -4y' = 4e^{2t}$$

$$\displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{-2t} + \frac{te^{2t}}{2} }$$

Problem Statement

Use the method of undetermined coefficients to find the general solution to $$y^{(3)} -4y' = 4e^{2t}$$

Solution

First, we solve the homogeneous equation $$y^{(3)} -4y' = 0$$.
$$\begin{array}{rcl} r^3 - 4r & = & 0 \\ r(r^2-4) & = & 0 \\ r = 0 & & r^2 = 4 \to r = \pm 2 \end{array}$$
$$y_h = A + Be^{2t} + Ce^{-2t}$$
Now we need to find the particular solution associated with $$4e^{2t}$$.
Since we already have an $$e^{2t}$$ factor in the homogeneous solution, we need to use $$y_p = Dte^{2t}$$, which includes the extra factor of $$t$$ to obtain an independent solution.

 $$y'_p = Dt(2e^{2t}) + De^{2t} = (2Dt+D)e^{2t}$$ $$y''_p = (2Dt+D)e^{2t}(2) + (2D)e^{2t} = (4Dt+4D)e^{2t}$$ $$y^{(3)}_p = (4Dt+4D)e^{2t}(2) + e^{2t}(4D) = (8Dt+12D)e^{2t}$$ $$y^{(3)} -4y' = 4e^{2t}$$ $$(8Dt+12D)e^{2t} - 4(2DT+D)e^{2t} = 4e^{2t}$$ $$(12D - 4D)e^{2t} = 4e^{2t}$$ $$8D = 4$$ $$D = 1/2$$ $$\displaystyle{ y_p = \frac{te^{2t}}{2} }$$

Combining the homogeneous and particular solutions gives us the general solution.

$$\displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{-2t} + \frac{te^{2t}}{2} }$$

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$$y^{(3)}-4y' = 5\cos(t)$$

Problem Statement

Use the method of undetermined coefficients to find the general solution to $$y^{(3)}-4y' = 5\cos(t)$$

$$y_p = -\sin(t)$$

Problem Statement

Use the method of undetermined coefficients to find the general solution to $$y^{(3)}-4y' = 5\cos(t)$$

Solution

We found the homogeneous solution in the previous problem. So here we need to find the particular solution. Looking at the right side of the equation, the form of our solution is $$y_p = A\cos(t)+B\sin(t)$$.

 $$y_p = A\cos(t)+B\sin(t)$$ $$y'_p=-A\sin(t)+B\cos(t)$$ $$y''_p=-A\cos(t)-B\sin(t)$$ $$y^{(3)}_p=A\cos(t)-B\cos(t)$$

Now plug in the derivatives above into the original differential equation to solve for A and B.
$$y^{(3)}-4y' = (A\cos(t)-B\cos(t))-4(-A\sin(t)+B\cos(t)) =$$ $$5A\sin(t)-5B\cos(t) = 5\cos(t))$$ $$\to A=0, B=-1$$

$$y_p = -\sin(t)$$

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Section 3

Solve the following problems.

Solve the initial value problem $$y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0$$

Problem Statement

Solve the initial value problem $$y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0$$

$$y = 9\cos(2t) - 4\cos(3t)$$

Problem Statement

Solve the initial value problem $$y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0$$

Solution

For this solution, we can apply the techniques of second order linear equations with constant coefficients. So our characteristic polynomial is $$r^4+13r^2+36=0 \to (r^2+4)(r^2+9)=0 \to r=\pm2i,\pm3i$$.
This gives us the general form of the solution as $$y = A\cos(2t) + B\sin(2t) + C\cos(3t) + D\sin(3t)$$.
Taking derivatives and solving for the coefficients, gives us $$A=9, B=0, C=-4, D=0$$.

$$y = 9\cos(2t) - 4\cos(3t)$$

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Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

Problem Statement

Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

$$y = Ae^t + Bte^t + Ct^2e^t + t^3e^t$$

Problem Statement

Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

Solution

This is a rather long problem but the 3x3 determinants reduce to manageable equations. We will show some intermediate results without all the details for calculating determinants.
For the homogeneous solution, we have constant coefficents, so our characteristic equation is $$r^3-3r^2+3r-1=0$$ $$\to$$ $$(r-1)^3=0$$. This gives us the homogeneous solution $$y_h=Ae^t+Bte^t+Ct^2e^t$$.
Now we need to set up the determinants for the variation of parameters method.

 $$y_1=e^t$$ $$y_2=te^t$$ $$y_3=t^2e^t$$

The Wronskian is
$$W = e^{3t} \begin{vmatrix} 1 & t & t^2 \\ 1 & t+1 & t^2+2t \\ 1 & t+2 & t^2+4t+2 \end{vmatrix}$$
This looks messy but it reduces to $$W=2e^{3t}$$. The variation of parameters technique gives us the following results.

 $$u'_1=3t^2 \to u_1=t^3$$ $$u'_2=-6t \to u_2=-3t^2$$ $$u'_3=3 \to u_3=3t$$

To get the particular solution, we use these results in the equation $$y_p = u_1 y_1 + u_2 y_2 + u_3 y_3 = t^3e^t$$ and our answer is $$y = y_h + y_p$$.

$$y = Ae^t + Bte^t + Ct^2e^t + t^3e^t$$

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You CAN Ace Differential Equations

 trig derivatives integration by parts separation of variables linear homogeneous wronskian undetermined coefficients variation of parameters linear systems

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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