## 17Calculus Differential Equations - Exam 4

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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IMPORTANT -
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Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2 hours

Questions

10

Total Points

175

Tools

Calculator

no

Formula Sheet(s)

provided

Other Tools

none

Instructions:
- 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).
- In the following panel, you will find the provided formulas.

### Provided Formulas

Bernoulli Equations: If $$\displaystyle{ \frac{dy}{dx} + p(x)y = q(x)y^n }$$ is transformed with the change of variables $$y=v^{1/(1-n)}$$, then the resulting equation is $$\displaystyle{ \frac{dv}{dx} + (1-n)p(x)v = (1-n)q(x) }$$ which is linear.
Homogeneous Equations: If $$\displaystyle{ \frac{dy}{dx} = F(y/x) }$$ is transformed with the change of variables $$y=vx$$, then the resulting equation is $$\displaystyle{ x\frac{dv}{dx} + v = F(v) }$$ which is separable.
Variation of Parameters: If $$y_1$$ and $$y_2$$ are linearly independent solutions of $$y''(t) + p(t)y'(t) + q(t)y = 0$$, then $Y(t) := -y_1(t) \int{ \frac{y_2(t)g(t)}{W(y_1,y_2)(t)} ~dt } + y_2(t) \int{ \frac{y_1(t)g(t)}{W(y_1,y_2)(t)} ~dt }$ is a solution of $$y''(t) + p(t)y'(t) + q(t)y = g(t)$$.
Euler Equations: The change of variables given by $$t=\ln x$$ transforms the equation $x^2y''(x) + \alpha x y'(x) + \beta y(x) = f(x)$ into the equation $y''(t) + (\alpha - 1)y'(t) + \beta y(t) = f(e^t).$ Regular Singular Points: If the equation $P(x)y''(x) + Q(x)y'(x) + R(x)y(x) = 0$ has a regular singular point at $$x_0$$ and we define $\alpha_0 := \lim_{x\to x_0}{ \frac{(x-x_0)\cdot Q(x)}{P(x)} } \text{ and } \beta_0 := \lim_{x\to x_0}{ \frac{(x-x_0)^2 \cdot R(x)}{P(x)} },$ then the indicial equation is given by $r^2 + (\alpha_0 - 1)r + \beta_0 .$ Classifying Linear Systems: If $$a, b, c$$ and $$d$$ are real numbers, and $x'(t) = ax(t) + by(t)$ $y'(t) = cx(t) + dy(t),$ then $$(0,0)$$ is an equilibrium point.
If we define $$D := ad - bc$$ and $$T := a + d$$, then we have the following classifications of the equilibrium point $$(0,0)$$.
(a) $$D \lt 0$$ → saddle
(b) $$0 \lt 4D \lt T^2$$ and
i. $$T \lt 0$$ → sink.
ii. $$T \gt 0$$ → source.
(c) $$T^2 \lt 4D$$ and
i. $$T \lt 0$$ → spiral sink.
ii. $$T = 0$$ → center.
iii. $$T \gt 0$$ → spiral source.

Function

Laplace Transform

$$f(t), g(t)$$

$$F(s), G(s)$$

$$1$$

$$\displaystyle{ \frac{1}{s}, ~ s \gt 0 }$$

$$e^{at}$$

$$\displaystyle{ \frac{1}{s-a}, ~ s \gt a }$$

$$t^n, ~ n \in \mathbb{N}; ~~ t^p, ~ p \gt -1$$

$$\displaystyle{ \frac{n!}{s^{n+1}}; ~~ \frac{\Gamma (p+1)}{s^{p+1}}, ~ s \gt 0 }$$

$$t^n e^{at}$$

$$\displaystyle{ \frac{n!}{(s-a)^{n+1}}, ~ s \gt a }$$

$$\sin(at), \cos(at)$$

$$\displaystyle{ \frac{a}{s^2+a^2}, ~ \frac{s}{s^2+a^2}, ~ s \gt 0 }$$

$$e^{at}\sin(bt), e^{at}\cos(bt)$$

$$\displaystyle{ \frac{b}{(s-a)^2+b^2}, \frac{s-a}{(s-a)^2+b^2}, ~ s \gt 0 }$$

$$u_c(t)$$

$$\displaystyle{ \frac{e^{-cs}}{s}, ~ s \gt 0 }$$

$$u_c(t)f(t-c)$$

$$e^{-cs}F(s)$$

$$e^{ct}f(t)$$

$$F(s-c)$$

$$\delta(t-c)$$

$$e^{-cs}$$

$$f(ct)$$

$$\displaystyle{ \frac{1}{c}F\left( \frac{s}{c} \right) }$$

$$\displaystyle{ f * g(t) = \int_0^t{ f(t-T)g(T)~dT } }$$

$$F(s)G(s)$$

Exam Questions

[20 points] Find the general solution of $y'(t) = -\frac{2}{t}y(t) + \frac{1}{t}e^{3t}$ for $$t \gt 0$$. (Do not use the Laplace Transform.)

Problem Statement

[20 points] Find the general solution of $y'(t) = -\frac{2}{t}y(t) + \frac{1}{t}e^{3t}$ for $$t \gt 0$$. (Do not use the Laplace Transform.)

$$\displaystyle{ y = \frac{1}{3t}e^{3t} + \frac{e^{3t}}{9t^2} + \frac{C}{t^2} }$$

Problem Statement

[20 points] Find the general solution of $y'(t) = -\frac{2}{t}y(t) + \frac{1}{t}e^{3t}$ for $$t \gt 0$$. (Do not use the Laplace Transform.)

Solution

First, we will rewrite this in a more standard form to help us categorize this equation.
$$\displaystyle{ y'(t) + \frac{2}{t}y(t) = \frac{1}{t}e^{3t} }$$
This is a first order differential equation in the form $$y' + p(t)y = g(t)$$, so we will use linear integrating factors.
Our integrating factor is $$\mu(t) = e^{\int{p(t)~dt}}$$ which can also be written $$\mu(t) = \exp{\int{p(t)~dt}}$$. We will use this last form to make the equations easier to read.
In our question, $$p(t)=2/t$$, so $$\mu(t) = \exp{\int{2/t~dt}} = \exp( 2\ln t) = t^2$$.
According to the linear integrating factors page, if we multiply both sides of the differential equation by the integrating factor, we should now have $$\displaystyle{ \frac{d}{dt}[t^2y] = te^{3t} }$$. Let's check to see if that is indeed the case.
$$\displaystyle{ \frac{d}{dt}[t^2y] = t^2y' + y(2t) }$$ ✓
Notice that we used the product rule here since $$y$$ is a function of $$t$$. So we are good to go.
Now we take $$\displaystyle{ \frac{d}{dt}[t^2y] = te^{3t} }$$ and integrate both sides.
$$\displaystyle{ \int{ \frac{d}{dt}[t^2y] ~dt } = \int{ te^{3t} ~dt } }$$
On the left, when we integrate the derivative, they cancel each other out, so we get $$t^2y$$.
On the right, we will use integration by parts.

 $$t^2y = \int{ te^{3t} ~dt }$$ $$u=t \to du = dt$$ and $$dv = e^{3t}dt \to v=e^{3t}/3$$ $$\displaystyle{ t^2y = \frac{t}{3}e^{3t} + \int{ \frac{e^{3t}}{3}~dt } }$$ $$\displaystyle{ t^2y = \frac{t}{3}e^{3t} + \frac{1}{3}\frac{e^{3t}}{3} + C }$$ $$\displaystyle{ y = \frac{1}{3t}e^{3t} + \frac{e^{3t}}{9t^2} + \frac{C}{t^2} }$$

We were specificially asked to find the general solution, which we have. We were not given initial conditions anyway. So the last line above is our final answer.

$$\displaystyle{ y = \frac{1}{3t}e^{3t} + \frac{e^{3t}}{9t^2} + \frac{C}{t^2} }$$

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[15 points] Write down the form of the best guess to use for the method of undetermined coefficients for the following ODE. $y'(t) + 9y'(t) + 14y(t) = 7\sin(3t) - 8e^{-2t} + 9e^{2t} - 10te^{-7t} + 11t^3 .$ You do not need to find the coefficients.

Problem Statement

[15 points] Write down the form of the best guess to use for the method of undetermined coefficients for the following ODE. $y'(t) + 9y'(t) + 14y(t) = 7\sin(3t) - 8e^{-2t} + 9e^{2t} - 10te^{-7t} + 11t^3 .$ You do not need to find the coefficients.

$$A\sin(3t) + B\cos(3t) + Ce^{2t} + Dt^3 + Et^2 + Ft + G + He^{-2t} + Ite^{-2t} + Je^{-7t} + Kte^{-7t} + Lt^2e^{-7t}$$

Problem Statement

[15 points] Write down the form of the best guess to use for the method of undetermined coefficients for the following ODE. $y'(t) + 9y'(t) + 14y(t) = 7\sin(3t) - 8e^{-2t} + 9e^{2t} - 10te^{-7t} + 11t^3 .$ You do not need to find the coefficients.

Solution

First, we need to find the solution to the homogeneous case, i.e. where $$y'(t) + 9y'(t) + 14y(t) = 0$$. This is easy since we have a second order linear equation with constant coefficients.
$$r^2 + 9r + 14 = 0 \to (r+7)(r+2)=0 \to r=-7, r=-2$$
$$y_h = c_1 e^{-7t} + c_2 e^{-2t}$$
So now we have to come up with terms associated with each of the factors on the right side of the equation. We will look at each term separately and then combine our answers at the end.

 $$7\sin(3t)$$ $$A\sin(3t) + B\cos(3t)$$ $$9e^{2t}$$ $$Ce^{2t}$$ $$11t^3$$ $$Dt^3 + Et^2 + Ft + G$$

Those were the easy ones since none of those terms appeared in the homogeneous solution. Now, let's look at the other terms.

 $$- 8e^{-2t}$$ $$He^{-2t} + Ite^{-2t}$$ $$- 10te^{-7t}$$ $$Je^{-7t} + Kte^{-7t} + Lt^2e^{-7t}$$

We were told not to find the coefficients. So our answer combines all the above terms.

$$A\sin(3t) + B\cos(3t) + Ce^{2t} + Dt^3 + Et^2 + Ft + G + He^{-2t} + Ite^{-2t} + Je^{-7t} + Kte^{-7t} + Lt^2e^{-7t}$$

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[15 points] Suppose you are given the following initial value problem. $y'(t) = 2t^2\sin(1+y'(t)^2 + y(t)^5), y(-2) = -3, y'(-2) = -5$ (a) Convert the problem above into a first order system.
(b) Set up the standard Euler iteration with a stepsize of 0.1. Explicitly give $$y_0$$ and $$z_0$$ and give $$y_{n+1}$$ and $$z_{n+1}$$ as functions of $$y_n$$, $$z_n$$ and $$n$$.
(c) In terms of this scheme, what would you use to approximate $$y(3)$$?

Problem Statement

[15 points] Suppose you are given the following initial value problem. $y'(t) = 2t^2\sin(1+y'(t)^2 + y(t)^5), y(-2) = -3, y'(-2) = -5$ (a) Convert the problem above into a first order system.
(b) Set up the standard Euler iteration with a stepsize of 0.1. Explicitly give $$y_0$$ and $$z_0$$ and give $$y_{n+1}$$ and $$z_{n+1}$$ as functions of $$y_n$$, $$z_n$$ and $$n$$.
(c) In terms of this scheme, what would you use to approximate $$y(3)$$?

Solution

(a) Let $$z = y'$$, so $$z' = y'$$. So we can write our first-order system.

$$\displaystyle{\begin{array}{rcl} y' & = & z \\ z' & = & 2t^2(1+z+y^5) \end{array} }$$

(b) We are given $$t_0 = -2$$ and $$h=0.1$$, so when $$n=0$$, $$y_0 = y(t_0) = y(-2) = -3$$ and $$z_0 = z(t_0) = z(-2) = y'(-2) = -5$$.
To see where we are going, let's set up $$n=1$$ and $$n=2$$.

 $$t_1 = t_0 + h$$ $$y_1 = y_0 + y'(t_0)h$$ $$z_1 = z_0 + z'(t_0)h$$
 $$t_2 = t_1 + h = t_0 + h + h = t_0 + 2h$$ $$y_2 = y_1 + y'(t_1)h$$ $$z_2 = z_1 + z'(t_1)h$$

From these we can extract the terms $$y_{n+1}$$ and $$z_{n+1}$$.
$$y_{n+1} = y_n + y'(t_n)h$$
$$z_{n+1} = z_n + z'(t_n)h$$ We just need to know what $$t_n$$ is in terms of $$n$$. From the above equations, we can extrapolate $$t_n = t_0 +nh = -2 + 0.1n$$

$$\displaystyle{\begin{array}{rcl} y_{n+1} & = & y_n + 0.1y'(0.1n-2) \\ z_{n+1} & = & z_n + 0.1z'(0.1n-2) \end{array} }$$

(c) To calculate $$y(3)$$, we need $$t_n=3$$, so $$0.1n-2 = 3 \to 0.1n = 5 \to n=50$$.

$$y_{51}$$

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[20 points] Solve using the the Laplace Transform. $x'(t) - 7x'(t) +10x(t) = 3\delta(t-6)- 5\delta(t-8)$ $x(0) = x'(0) = 0$

Problem Statement

[20 points] Solve using the the Laplace Transform. $x'(t) - 7x'(t) +10x(t) = 3\delta(t-6)- 5\delta(t-8)$ $x(0) = x'(0) = 0$

$$\displaystyle{ x(t) = u_6(t)e^{5(t-6)} - u_6(t)e^{2(t-6)} - (5/3) u_8(t)e^{5(t-8)} + (5/3) u_8(t)e^{2(t-8)} }$$

Problem Statement

[20 points] Solve using the the Laplace Transform. $x'(t) - 7x'(t) +10x(t) = 3\delta(t-6)- 5\delta(t-8)$ $x(0) = x'(0) = 0$

Solution

Here are some Laplace Transform rules we will use.

 $$\mathcal{L}\{ x(t) \} = X(s)$$ $$\mathcal{L}\{ x'(t) \} = sX(s) - x(0)$$ $$\mathcal{L}\{ x'(t) \} = s^2X(s) - sx(0) - x'(0)$$ $$\mathcal{L}\{ \delta(t-c) \} = e^{-cs}$$

Now apply the Laplace Transform to the differential equation.

 $$\mathcal{L}\{ x'(t) - 7x'(t) +10x(t) \} = \mathcal{L}\{ 3\delta(t-6)- 5\delta(t-8) \}$$ $$s^2X(s) - 7sX(s) + 10X(s) = 3e^{-6s} - 5e^{-8s}$$ $$X(s)( s^2 - 7s + 10 ) = 3e^{-6s} - 5e^{-8s}$$ $$\displaystyle{ X(s) = \frac{3e^{-6s}-5e^{-8s}}{(s-5)(s-2)} }$$

Since the denominator factored, this gives us the idea to expand into partial fractions with the result
$$\displaystyle{ \frac{1}{(s-5)(s-2)} = \frac{1/3}{s-5} + \frac{-1/3}{s-2} }$$

 $$\displaystyle{ X(s) = \left[ 3e^{-6s} - 5e^{-8s} \right] \left[ \frac{1/3}{s-5} - \frac{1/3}{s-2} \right] }$$ $$\displaystyle{ X(s) = \frac{e^{-6s}}{s-5} - \frac{e^{-6s}}{s-2} - \frac{(5/3)e^{-8s}}{s-5} + \frac{(5/3)e^{-8s}}{s-2} }$$

Now each fraction is in a form where we can take the inverse Laplace Transform.
$$\displaystyle{ \frac{1}{s-5} \to e^{5t} }$$ and $$\displaystyle{ \frac{1}{s-2} \to e^{2t} }$$
$$e^{6s}F(s) \to u_6(t)F(t-6)$$ and $$e^{8s}F(s) \to u_8(t)F(t-8)$$
Term by term, we have the following.

 $$\displaystyle{ \frac{e^{-6s}}{s-5} \to u_6(t)e^{5(t-6)} }$$ $$\displaystyle{ \frac{e^{-6s}}{s-2} \to u_6(t)e^{2(t-6)} }$$ $$\displaystyle{ \frac{e^{-8s}}{s-5} \to u_8(t)e^{5(t-8)} }$$ $$\displaystyle{ \frac{e^{-8s}}{s-2} \to u_8(t)e^{2(t-8)} }$$

Our final answer is the sum of these terms with the appropriate coefficients.
Note: There are more compact ways to write this answer. However, since this is a exam question, your instructor may accept the answer in this form. As usual, check with your instructor to see what they expect.

$$\displaystyle{ x(t) = u_6(t)e^{5(t-6)} - u_6(t)e^{2(t-6)} - (5/3) u_8(t)e^{5(t-8)} + (5/3) u_8(t)e^{2(t-8)} }$$

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[15 points] Find $$y(t)$$ using the Laplace Transform. (You do not need to find $$x(t)$$.) $\begin{array}{ll} x'(t) = x(t) + 3y(t) & x(0) = 1 \\ y'(t) = -3x(t) + y(t) & y(0) = 2 \end{array}$

Problem Statement

[15 points] Find $$y(t)$$ using the Laplace Transform. (You do not need to find $$x(t)$$.) $\begin{array}{ll} x'(t) = x(t) + 3y(t) & x(0) = 1 \\ y'(t) = -3x(t) + y(t) & y(0) = 2 \end{array}$

$$y(t) = e^t[2\cos(3t) - \sin(3t)]$$

Problem Statement

[15 points] Find $$y(t)$$ using the Laplace Transform. (You do not need to find $$x(t)$$.) $\begin{array}{ll} x'(t) = x(t) + 3y(t) & x(0) = 1 \\ y'(t) = -3x(t) + y(t) & y(0) = 2 \end{array}$

Solution

We will use the Laplace Transform identity $$\mathcal{L}\{ y'(t) \} = sY(s) - y(0)$$. A similar identity applies to $$x'(t)$$.

 $$\mathcal{L}\{ y'(t) = -3x(t) + y(t) \}$$ where $$y(0) = 2$$ $$sY(s) - 2 = -3X(s) + Y(s)$$ $$sY(s) - Y(s) = 2 - 3X(s)$$ $$Y(s)(s-1) = 2-3X(s)$$ Now we apply the same technique to the first equation. $$\mathcal{L}\{ x'(t) = x(t) + 3y(t) \}$$ where $$x(0) = 1$$ $$sX(s) - 1 = X(s) + 3Y(s)$$ Since we are asked to solve for $$y(t)$$, we will solve the second equation for $$X(s)$$ and substitute it into the first equation. $$X(s)(s-1) = 3Y(s) + 1$$ $$\displaystyle{ X(s) = \frac{3Y(s)+1}{s-1} }$$ Substitute back into the first equation. $$\displaystyle{ Y(s)(s-1) = 2-3\frac{3Y(s)+1}{s-1} }$$ $$Y(s)(s-1)^2 = 2(s-1)-3(3Y(s)+1)$$ $$Y(s)(s-1)^2 = 2(s-1)-9Y(s)-3$$ $$Y(s)(s-1)^2 + 9Y(s) = 2(s-1)-3$$ $$\displaystyle{ Y(s) = \frac{2(s-1)-3}{(s-1)^2+9} }$$ Now we need to convert back to the time domain. We will separate the expression on the right into two fractions so that it easier to see what we have to work with. $$\displaystyle{ Y(s) = \frac{2(s-1)}{(s-1)^2+9} - \frac{3}{(s-1)^2+9} }$$ Now we will use the following identities to convert back to the time domain. $$\displaystyle{ \mathcal{L}\{ e^{at}\sin(bt) \} = \frac{b}{(s-a)^2+b} }$$ $$\displaystyle{ \mathcal{L}\{ e^{at}\cos(bt) \} = \frac{s-a}{(s-a)^2+b} }$$ $$\displaystyle{ \mathcal{L^{-1}}\left\{ \frac{3}{(s-1)^2+9} \right\} = e^t\sin(3t) }$$ $$\displaystyle{ \mathcal{L^{-1}}\left\{ \frac{s-1}{(s-1)^2+9} \right\} = e^t\cos(3t) }$$ Combining the last two results, we get our final answer.

$$y(t) = e^t[2\cos(3t) - \sin(3t)]$$

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[12 points] Solve $y'(x) + 2xy'(x) + y(x) = 0 ~~~ y(0) = 0, y'(0) = 3$ by using a power series centered at zero. Be sure to write down a recurrence relation and find the first three nonzero terms in the expansion.

Problem Statement

[12 points] Solve $y'(x) + 2xy'(x) + y(x) = 0 ~~~ y(0) = 0, y'(0) = 3$ by using a power series centered at zero. Be sure to write down a recurrence relation and find the first three nonzero terms in the expansion.

Solution

For a power series, our answer will be in the form $$\displaystyle{ y(x) = \sum_{n=0}^{\infty}{ a_n x^n } }$$.
From this, we take the first and second derivatives and plug them back into the differential equation.

 $$\displaystyle{ y(x) = \sum_{n=0}^{\infty}{ a_n x^n } }$$ $$\displaystyle{ y'(x) = \sum_{n=1}^{\infty}{ n a_n x^{n-1} } }$$ $$\displaystyle{ y'(x) = \sum_{n=2}^{\infty}{ n(n-1)a_n x^{n-2} } }$$ Now plug these derivatives into the differential equation. $$\displaystyle{ \sum_{n=2}^{\infty}{ n(n-1)a_n x^{n-2} } + 2\sum_{n=1}^{\infty}{ n a_n x^{n} } + \sum_{n=0}^{\infty}{ a_n x^n } = 0 }$$

Before we can combine the sums we need to do two things. First, we need to get all the exponents on the x's to be the same. This will allow us to factor it out. We choose to get them all to be $$x^n$$. This means that we need to replace n with n+2 in the first sum. When we do that we get the first term as
$$\displaystyle{ \sum_{n=0}^{\infty}{ (n+2)(n+1)a_{n+2} x^{n} } }$$

Secondly, we need to get the indices on the sums to start at the same value. We choose to have them all start at $$n=1$$. So we just need to pull out the first term from the first and third series. Our result is
$$\displaystyle{ 2a_2 + \sum_{n=1}^{\infty}{ (n+2)(n+1)a_{n+2} x^{n} } + 2\sum_{n=1}^{\infty}{ n a_n x^{n} } + a_0 + \sum_{n=1}^{\infty}{ a_n x^n } = 0 }$$
Now we combine the sums and move the constants together to get
$$\displaystyle{ (a_0 + 2a_2) + \sum_{n=1}^{\infty}{ \left[ (n+2)(n+1)a_{n+2} + 2na_n + a_n \right]x^n = 0 } }$$

If we stand back and look at this, we notice that the right side is zero. This means that all coefficients on the left also have to be zero. So now we have
$$\displaystyle{ a_0 + 2a_2 = 0 \to a_2 = \frac{-a_0}{2} }$$ and
$$\displaystyle{ (n+2)(n+1)a_{n+2} + 2na_n + a_n = 0 \to a_{n+2} = \frac{-(2n+1)a_n}{(n+2)(n+1)} }$$
Now we can start to build some coefficients in order to find a pattern.

 $$n=0$$
 $$\displaystyle{ a_2 = \frac{-a_0}{2} }$$
 $$n=1$$
 $$\displaystyle{ a_3 = \frac{-3a_1}{2 \cdot 3} }$$
 $$n=2$$
 $$\displaystyle{ a_4 = \frac{-5a_2}{3 \cdot 4} = \frac{-5}{3 \cdot 4}\frac{-a_0}{2} = \frac{5a_0}{4!} }$$
 $$n=3$$
 $$\displaystyle{ a_5 = \frac{-7a_3}{4 \cdot 5} = \frac{-7}{4 \cdot 5}\frac{-3a_1}{2 \cdot 3} = \frac{3 \cdot 7 a_1}{5!} }$$
 $$n=4$$
 $$\displaystyle{ a_6 = \frac{-9a_4}{5 \cdot 6} = \frac{-9}{5 \cdot 6}\frac{5a_0}{4!} = \frac{-5 \cdot 9 a_0}{6!} }$$
 $$n=5$$
 $$\displaystyle{ a_7 = \frac{-11a_5}{6 \cdot 7} = \frac{-11}{6 \cdot 7}\frac{7 \cdot 3a_1}{5!} = \frac{-11 \cdot 7 \cdot 3a_1}{7!} }$$
 $$n=6$$
 $$\displaystyle{ a_8 = \frac{-13a_6}{7 \cdot 8} = \frac{-13}{7 \cdot 8}\frac{-9 \cdot 5a_0}{6!} = \frac{13 \cdot 9 \cdot 5 a_0}{8!} }$$

Okay, so now let's build $$y(x)$$. We will change the index variable on the sum to something other than n to avoid confusion.
$$\displaystyle{ y(x) = \sum_{k=0}^{\infty}{ a_k x^k } = a_0 + a_1 x - \frac{a_0}{2}x^2 - \frac{3a_1}{3!}x^3 + \frac{5a_0}{4!}x^4 + \frac{3 \cdot 7a^1}{5!}x^5 - \frac{5 \cdot 9 a_0}{6!}x^6 - \frac{3 \cdot 7 \cdot 11 a_1}{7!}x^7 + \frac{5 \cdot 9 \cdot 13}{8!}x^8 + . . . }$$
Using the initial conditions, we can determine $$a_0 = 0$$ and $$a_1 = 3$$. So now we have
$$\displaystyle{ y(x) = 3x - \frac{3}{3!}(3x^3) + \frac{3 \cdot 7}{5!}(3x^5) - \frac{3 \cdot 7 \cdot 11}{7!}(3x^3) + . . . }$$
We use a third index variable to keep things straight.

$$\displaystyle{y(x) = 3x + 3\sum_{m=1}^{\infty}{ \left[ \frac{ (-1)^m [3 \cdot 7 \cdot 11 \cdot \cdot \cdot (4(m-1)+3)] x^{2m+1} }{(2m+1)!} \right] } }$$

The first 3 nonzero terms are

$$\displaystyle{3x - \frac{3}{2}x^3 + \frac{21}{40}x^5 }$$

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[18 points] Consider the differential equation $x^3(x+1) y' + 3x^2 y' + y = 0.$ Zero and $$-1$$ are obviously singular points.
(a) Classify the singular points as regular or irregular.
(b) At any regular singular points, find the indicial equation and the roots of the indicial equation.
(c) At any regular singular point, state where the series solution corresponding to the larger root would converge.
Do not try to find the solution. You do not have time.

Problem Statement

[18 points] Consider the differential equation $x^3(x+1) y' + 3x^2 y' + y = 0.$ Zero and $$-1$$ are obviously singular points.
(a) Classify the singular points as regular or irregular.
(b) At any regular singular points, find the indicial equation and the roots of the indicial equation.
(c) At any regular singular point, state where the series solution corresponding to the larger root would converge.
Do not try to find the solution. You do not have time.

(a) $$x=0$$ is irregular; $$x=-1$$ is regular.
(b) $$(r+1)^2$$; $$r=-1$$, $$r=-1$$
(c) $$(-1,0)$$

Problem Statement

[18 points] Consider the differential equation $x^3(x+1) y' + 3x^2 y' + y = 0.$ Zero and $$-1$$ are obviously singular points.
(a) Classify the singular points as regular or irregular.
(b) At any regular singular points, find the indicial equation and the roots of the indicial equation.
(c) At any regular singular point, state where the series solution corresponding to the larger root would converge.
Do not try to find the solution. You do not have time.

Solution

(a) In order to classify the singular points as regular or irregular, we need to calculate the following limits for each singular point, $$x_0$$. $\lim_{x \to x_0}{ (x-x_0) \frac{Q(x)}{P(x)} } \text{ and } \lim_{x \to x_0}{ (x-x_0)^2 \frac{Q(x)}{R(x)} }$ where $$P(x) = x^3(x+1)$$, $$Q(x) = 3x^2$$ and $$R(x) = 1$$, taken from the coefficients of $$y'$$, $$y'$$ and $$y$$, respectively.

 First, let's look at $$x=0$$. $$\displaystyle{ \lim_{x \to 0}{ x\frac{3x^2}{x^3(x+1)} } }$$ $$\displaystyle{ \lim_{x \to 0}{ \frac{3x^3}{x^3(x+1)} } }$$ $$\displaystyle{ \lim_{x \to 0}{ \frac{3}{x+1} } = 3 }$$ Since the limit is finite, we will now check the second limit. $$\displaystyle{ \lim_{x \to 0}{ x^2\frac{1}{x^3(x+1)} } }$$ $$\displaystyle{ \lim_{x \to 0}{ \frac{1}{x(x+1)} } = \frac{1}{0} = \infty }$$ Since the second limit is not finite, the singular point $$x=0$$ is irregular. Now let's look at $$x=-1$$ $$\displaystyle{ \lim_{x \to -1}{ (x+1)\frac{3x^2}{x^3(x+1)} } }$$ $$\displaystyle{ \lim_{x \to -1}{ \frac{3}{x} } = \frac{3}{-1} = -3 }$$ Since this limit is finite, we will need to check the second limit. $$\displaystyle{ \lim_{x \to -1}{ (x+1)^2 \frac{1}{x^3(x+1)} } }$$ $$\displaystyle{ \lim_{x \to -1}{ \frac{x+1}{x^3} } = \frac{0}{-1} = 0 }$$ Since both limits are finite, $$x=-1$$ is a regular singular point.

(b) For this part of the problem, we need to rewrite the differential equation in the form $y' + \frac{Q(x)}{P(x)}y' + \frac{R(x)}{P(x)}y = 0$ So we get $y' + \frac{3x^2}{x^3(x+1)}y' + \frac{1}{x^3(x+1)} = 0$ $y' + \frac{3}{x(x+1)}y' + \frac{1}{x^3(x+1)} = 0$ The equation for the indicial equation is $$r^2 + (\alpha_0-1)r + \beta_0 = 0$$. We need to determine $$\alpha_0$$ and $$\beta_0$$.
To get $$\alpha_0$$ we need to look at $$\displaystyle{ (x+1)\frac{Q(x)}{P(x)} = (x+1)\frac{3}{x(x+1)} = \frac{3}{x} = 3x^{-1} }$$
$$\alpha_0$$ is the coefficent of the lowest term, so $$\alpha_0=3$$.
To get $$\beta_0$$ we need to look at $$\displaystyle{ (x+1)^2 \frac{R(x)}{P(x)} = \frac{x+1}{x^3} = x^{-2} + x^{-3} }$$
$$\beta_0$$ is the coefficient of the lowest term, so $$\beta_0=1$$
Now we have $$r^2 + (\alpha_0-1)r + \beta_0 = 0 \to r^2 + 2r + 1 = 0 \to (r+1)^2 = 0$$

(c) This part is easy now, since we know that the radius of convergence is $$0 \leq x+1 \leq R$$ where R is the distance to the next closest singular point from the largest. In our case the distance is R=1, so we have
$$0 \lt x+1 \lt 1 \to -1 \lt x \lt 0$$

(a) $$x=0$$ is irregular; $$x=-1$$ is regular.
(b) $$(r+1)^2$$; $$r=-1$$, $$r=-1$$
(c) $$(-1,0)$$

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[20 points] The equation $x^2 y'(x) + (x^2 + x)y'(x) - 4y(x) = 0$ has a regular singular point at zero. For this equation, $$\alpha_0 = 1$$ and $$\beta_0 = -4$$, and so the indicial equation is $$s^2-4=0$$. Find a series solution to the ODE above which corresponds to the larger root of the indicial equation.

Problem Statement

[20 points] The equation $x^2 y'(x) + (x^2 + x)y'(x) - 4y(x) = 0$ has a regular singular point at zero. For this equation, $$\alpha_0 = 1$$ and $$\beta_0 = -4$$, and so the indicial equation is $$s^2-4=0$$. Find a series solution to the ODE above which corresponds to the larger root of the indicial equation.

$$\displaystyle{ y = a_0 + \sum_{n=1}^{\infty}{ \frac{24a_0(n+1)}{(n+4)!} x^n } }$$

Problem Statement

[20 points] The equation $x^2 y'(x) + (x^2 + x)y'(x) - 4y(x) = 0$ has a regular singular point at zero. For this equation, $$\alpha_0 = 1$$ and $$\beta_0 = -4$$, and so the indicial equation is $$s^2-4=0$$. Find a series solution to the ODE above which corresponds to the larger root of the indicial equation.

Solution

The larger root of $$s^2-4=0$$ is $$s=2$$. So our series solution is $$\displaystyle{y=x^2 \sum_{n=0}^{\infty}{ a_n x^n } = \sum_{n=0}^{\infty}{ a_n x^{n+2} } }$$
Let's take the derivatives of y to prepare for putting them into the differential equation.

 $$\displaystyle{ y' = \sum_{n=0}^{\infty}{ a_n (n+2) x^{n+1} } }$$ $$\displaystyle{ y' = \sum_{n=0}^{\infty}{ a_n (n+2)(n+1)x^n } }$$ Put the series into the differential equation $$x^2 y'(x) + (x^2 + x)y'(x) - 4y(x) = 0$$ $$\displaystyle{x^2 \sum_{n=0}^{\infty}{ a_n (n+2)(n+1)x^n } + (x^2 + x) \sum_{n=0}^{\infty}{ a_n (n+2) x^{n+1} } - 4 \sum_{n=0}^{\infty}{ a_n x^{n+2} } = 0 }$$ Move the terms outside the sums into the sums and simplify. $$\displaystyle{\sum_{n=0}^{\infty}{ a_n (n+2)(n+1)x^{n+2} } + \sum_{n=0}^{\infty}{ a_n (n+2) x^{n+3} } + \sum_{n=0}^{\infty}{ a_n (n+2) x^{n+2} } - 4 \sum_{n=0}^{\infty}{ a_n x^{n+2} } = 0 }$$ Get all the exponents on x to be the same. We will change only the second term to get $$x^{n+2}$$. Replacing n with n-1 changes the second term to $$\displaystyle{\sum_{n=1}^{\infty}{ a_{n-1} (n+1) x^{n+2} } }$$ Now we will remove the first term in the other three sums to that all sums start at n=1. $$\displaystyle{2a_0 x^2 + \sum_{n=1}^{\infty}{ a_n (n+2)(n+1)x^{n+2} } + \sum_{n=1}^{\infty}{ a_{n-1} (n+1) x^{n+2} } + 2a_0 x^2 + \sum_{n=1}^{\infty}{ a_n (n+2) x^{n+2} } - 4a_0 x^2 - 4 \sum_{n=1}^{\infty}{ a_n x^{n+2} } = 0 }$$ As expected, the outside terms cancel, i.e. $$(2a_0 + 2a_0 - 4a_0) x^2 = 0$$ and the sums are the correct form to combine to get $$\displaystyle{\sum_{n=1}^{\infty}{ [ a_n(n+2)(n+1) + a_{n-1}(n+1) + a_n(n+2) - 4a_n ]x^{n+2} } = 0 }$$ Since all coefficients should be zero because the right side is zero, we have $$a_n(n+2)(n+1) + a_{n-1}(n+1) + a_n(n+2) - 4a_n = 0$$ Solving for $$a_n$$ and simplifying, we have $$\displaystyle{ a_n = \frac{-a_{n-1}(n+1)}{n(n+4)} }$$

Now we will build a table to see if we can find a pattern for $$a_n$$.

 $$n=1$$ $$\displaystyle{ a_1 = \frac{-2a_0}{1 \cdot 5} }$$ $$n=2$$ $$\displaystyle{ a_2 = \frac{-3a_1}{2 \cdot 6} = \frac{2 \cdot 3 a_0}{(2\cdot6)(1\cdot5)} }$$ $$n=3$$ $$\displaystyle{ a_3 = \frac{-4a_2}{3\cdot 7} = \frac{-2\cdot 3\cdot 4 a_0}{(3\cdot7)(2\cdot6)(1\cdot5)} }$$ $$n=4$$ $$\displaystyle{ a_4 = \frac{-5a_3}{4\cdot 8} = \frac{2\cdot3\cdot4\cdot5a_0}{(4\cdot 8)(3\cdot7)(2\cdot6)(1\cdot5)} }$$

From these terms we can see that the pattern is $$\displaystyle{ a_n = \frac{24a_0 (n+1)!}{n!(n+4)!} = \frac{24a_0(n+1)}{(n+4)!} }$$

$$\displaystyle{ y = a_0 + \sum_{n=1}^{\infty}{ \frac{24a_0(n+1)}{(n+4)!} x^n } }$$

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[20 points] (a) Write down the definition of the Laplace Transform $$\mathcal{L}[f]$$.
(b) Find all solutions of $y'(x) + 4y(x) = 0; y(0) = -1, y(\pi/2) = 1$ (c) Using the fact that $$x^{-1}$$ and $$x^{-2}$$ are linearly independent solutions of $x^2y'(x) + 4xy'(x) + 2y(x) = 0,$ find the general solution of the following problem for $$x \gt 0$$ $x^2y'(x) + 4xy'(x) + 2y(x) = e^{2x}$

Problem Statement

[20 points] (a) Write down the definition of the Laplace Transform $$\mathcal{L}[f]$$.
(b) Find all solutions of $y'(x) + 4y(x) = 0; y(0) = -1, y(\pi/2) = 1$ (c) Using the fact that $$x^{-1}$$ and $$x^{-2}$$ are linearly independent solutions of $x^2y'(x) + 4xy'(x) + 2y(x) = 0,$ find the general solution of the following problem for $$x \gt 0$$ $x^2y'(x) + 4xy'(x) + 2y(x) = e^{2x}$

(a) $$\displaystyle{ \mathcal{L}[ f(t) ] = F(s) = \int_{0}^{\infty}{e^{-st}f(t)~dt} }$$
(b) $$y=-\cos(2x)+B\sin(2x)$$
(c)
$y(x) = \frac{c_1}{x} + \frac{c_2}{x^2} + \left[ \frac{x}{2} - \frac{1}{2} + \frac{1}{4x} \right] e^{2x} + \left[ \frac{-x}{3} + \frac{3}{4} - \frac{3}{4x} + \frac{3}{8x^2} \right] e^{2x}$

Problem Statement

[20 points] (a) Write down the definition of the Laplace Transform $$\mathcal{L}[f]$$.
(b) Find all solutions of $y'(x) + 4y(x) = 0; y(0) = -1, y(\pi/2) = 1$ (c) Using the fact that $$x^{-1}$$ and $$x^{-2}$$ are linearly independent solutions of $x^2y'(x) + 4xy'(x) + 2y(x) = 0,$ find the general solution of the following problem for $$x \gt 0$$ $x^2y'(x) + 4xy'(x) + 2y(x) = e^{2x}$

Solution

(a) $$\displaystyle{ \mathcal{L}[ f(t) ] = F(s) = \int_{0}^{\infty}{e^{-st}f(t)~dt} }$$

(b) Although not required in the problem statement, we assume that the instructor wants us to use the Laplace Transform to answer this part. But after looking at the given data, we are not given $$y'(0)$$ which the Laplace Transform requires for $$\mathcal{L}[y']$$. So we will use the technique for second order linear with constant coefficients.
Our characteristic equation is $$r^2+4=0 \to r = \pm 2i$$. So our general solution is $$y=A\cos(2x) + B\sin(2x)$$. Using the initial conditions gives us $$A=-1$$ for both. We are not given enough information to determine B. Since they want all solutions, our final answer is $$y=-\cos(2x)+B\sin(2x)$$.

(c) Again, we are not told which technique to use, so we choose to use variation of parameters since we are given the homogeneous solutions.

 $$y_1 = x^{-1} \to y_1' = -1x^{-2}$$ $$y_2 = x^{-2} \to y_2' = -2x^{-3}$$ Now we calculate the Wronskian. $$W = \begin{vmatrix} x^{-1} & x^{-2} \\ -1x^{-2} & -2x^{-3} \end{vmatrix} = -x^{-4}$$ $$\displaystyle{u_1' = \frac{\begin{vmatrix} 0 & x^{-2} \\ e^{2x} & -2x^{-3} \end{vmatrix}}{W} = x^2 e^{2x} }$$ $$\displaystyle{u_2' = \frac{\begin{vmatrix} x^{-1} & 0 \\ -1x^{-2} & e^{2x} \end{vmatrix} }{W} = -x^3 e^{2x} }$$

Just to clarify where we are going, our solution will be $$y(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$$. So now we need to integrate $$u_1'$$ and $$u_2'$$.
$$u_1 = \int{ x^2 e^{2x}~dx }$$ and $$u_2 = \int{ -x^3 e^{2x} ~dx }$$
Both of these integrals require the integration by parts. The results give us the following general solution. $y(x) = \frac{c_1}{x} + \frac{c_2}{x^2} + \left[ \frac{x}{2} - \frac{1}{2} + \frac{1}{4x} \right] e^{2x} + \left[ \frac{-x}{3} + \frac{3}{4} - \frac{3}{4x} + \frac{3}{8x^2} \right] e^{2x}$ Note - Some simplification and combining terms in this answer can be done and would usually be required on homework but your instructor may allow this as a final answer for an exam. Check with them to see what they require.

(a) $$\displaystyle{ \mathcal{L}[ f(t) ] = F(s) = \int_{0}^{\infty}{e^{-st}f(t)~dt} }$$
(b) $$y=-\cos(2x)+B\sin(2x)$$
(c)
$y(x) = \frac{c_1}{x} + \frac{c_2}{x^2} + \left[ \frac{x}{2} - \frac{1}{2} + \frac{1}{4x} \right] e^{2x} + \left[ \frac{-x}{3} + \frac{3}{4} - \frac{3}{4x} + \frac{3}{8x^2} \right] e^{2x}$

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[20 points] (a) Find the general solution of the following ODE. $y'(t) + 4y(t) = 4t^2$ (b) For the autonomous differential equation $y'(t) = (y+2)(y-7)$ determine the equilibrium points and classify them as stable, semistable or unstable.

Problem Statement

[20 points] (a) Find the general solution of the following ODE. $y'(t) + 4y(t) = 4t^2$ (b) For the autonomous differential equation $y'(t) = (y+2)(y-7)$ determine the equilibrium points and classify them as stable, semistable or unstable.

(a) $$y(t) = c_1 \cos(2t) + c_2 \sin(2t) -1/2 + t^2$$
(b) $$y=-2$$ is stable and $$y=7$$ is semi-stable.

Problem Statement

[20 points] (a) Find the general solution of the following ODE. $y'(t) + 4y(t) = 4t^2$ (b) For the autonomous differential equation $y'(t) = (y+2)(y-7)$ determine the equilibrium points and classify them as stable, semistable or unstable.

Solution

(a) First we will solve the homogeneous equation $$y'(t) + 4y(t) = 0$$. This is a simple second-order equation with constant coefficients. So we have the characteristic equation $$r^2+4=0 \to r=\pm 2i$$ giving us the homogeneous solution $$y_h(t) = c_1 \cos(2t) + c_2 \sin(2t)$$.
To find the inhomogenous solution, we will use undetermined coefficients based on $$g(t) = 4t^2$$.

 $$y_p(t) = A+Bt+Ct^2$$ $$y_p'(t) = B+2Ct$$ $$y_p'(t) = 2C$$ $$y'(t) + 4y(t) = 4t^2$$ $$2C + 4(A+Bt+Ct^2) = 4t^2$$

Solving for the constants gives us $$A=-1/2, B=0, C=1$$.
So the particular solution is $$y_p(t) = -1/2 + t^2$$. The general solution is the combination of the homogeneous and particular solutions, i.e. $$y(t) = y_h(t) + y_p(t)$$.

(b) The equilibrium points are where $$y'(t) = 0$$, so we have $$(y+2)(y-7) = 0 \to y=-2 \text{ and } y=7$$.
If we build a table, we can see that for values of $$y \gt 7$$ the slope is positive and therefore diverging up to infinity. So the equilibrium point $$y = 7$$ is either unstable or semi-stable, depending what the slope does between $$y=-2$$ and $$y = 7$$.
Let's look at values of $$y \lt -2$$ first. For these values, the slope is positive, which means functions below $$y = -2$$ tend toward $$y = -2$$.
Between the two values, the slope is negative, which means the graphs tend toward $$y = -2$$.
Putting all this together, we have $$y=-2$$ stable and $$y=7$$ semi-stable.

(a) $$y(t) = c_1 \cos(2t) + c_2 \sin(2t) -1/2 + t^2$$
(b) $$y=-2$$ is stable and $$y=7$$ is semi-stable.

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You CAN Ace Differential Equations

 precalculus - partial fractions calculus - integration by parts linear homogeneous wronskian variation of parameters linear integrating factors laplace transforms power series solution singular points first order systems Euler's Method

### 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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