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Differential Equations : Differential Equations

Study concepts, example questions & explanations for Differential Equations

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Example Questions

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Example Question #1 : Euler Method

Use Euler's Method to calculate the approximation of y(0.2) where y(x) is the solution of the initial-value problem that is as follows.

y''+xy'+y=0

y(0)=2, y'(0)=3

Possible Answers:

$y(0.2)\approx2.58$

$y(0.2)\approx2.584$

$y(0.2)\approx2.542$

$y(0.2)\approx2.458$

$y(0.2)\approx2.5$

Correct answer:

$y(0.2)\approx2.58$

Explanation:

Using Euler's Method for the function

y''+xy'+y=0

y(0)=2, y'(0)=3

first make the substitution of

$\\y'=u \\u'=-xu-y$

therefore

$\\y_{n+1}=y_n+hu_n \\u_{n+1}=u_n+h[-x_nu_n-y_n]$

where represents the step size.

Let

$\\h=0.1 \\y_0=2 \\u_0=3$

Substitute these values into the previous formulas and continue in this fashion until the approximation for y(0.2) is found.

$\\y_1=y_0+(0.1)u_0=2+(0.1)(3)=2.3 \\u_1=u_0+(0.1)[-x_0u_0-y_0]=3+(0.1)[0(2)-2]=2.8 \\y_2=y_1+(0.1)u_1=2.3+(0.1)(2.8)=2.58 \\u_2=u_1+(0.1)[-x_1u_1-y_1]=2.8+(0.1)[-0.1(2.8)-2.3]=2.542$

Therefore,

$y(0.2)\approx2.58$

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Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Use the implicit Euler method to approximate y(3) for y' = 5y , given that y(0) = 4 , using a time step of h = 1.

Possible Answers:

$-\frac{1}{16}$

$\frac{1}{16}$

$\frac{1}{32}$

Correct answer:

$-\frac{1}{16}$

Explanation:

In the implicit method, the amount to increase is given by $h \cdot f(t_{i+1}, y_{i+1})$ , or in this case $\mu_i = \mu_{i-1}+ 1 \cdot (5\mu_i)$ . Note, you can't just plug in to this form of the equation, because it's implicit: $\mu_i$ is on both sides. Thankfully, this is an easy enough form that you can solve explicitly. Otherwise, you would have to use an approximation method like newton's method to find $\mu_i$ . Solving explicitly, we have $-4\mu_i = \mu_{i-1}$ and $\mu_i = -\frac{\mu_{i-1}}{4}$ .

Thus, $y(1) \approx \mu_1 = -\frac{4}{4} = -1$

$y(2) \approx \mu_2 = -\frac{-1}{4} = \frac{1}{4}$

$y(3) \approx \mu_3 = -\frac{\frac{1}{4}}{4} = -\frac{1}{16}$

Thus, we have a final answer of $-\frac{1}{16}$

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Example Question #1 : Euler Method

Use two steps of Euler's Method with h=0.1 on

$y' = x\sqrt{y} \ \ \ y(1) = 4$

To three decimal places

Possible Answers:

4.425

4.420

4.413

4.428

4.408

Correct answer:

4.425

Explanation:

Euler's Method gives us

$y_{n+1} = y_n + hf(x_n,y_n) = y_n + (0.1)x_n\sqrt{y_n}$

Taking one step

$y_1 = y_0 + (0.1)x_0\sqrt{y_0} = 4 + (0.1)(1)\sqrt{4} = 4.2$

Taking another step

$y_2 = y_1 + (0.1)x_1\sqrt{y_1} = 4.2 + (0.1)(1.1)\sqrt{4.2} \approx 4.425$

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Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

The two-step Adams-Bashforth method of approximation uses the approximation scheme $y_{i+2} = y_{i+1} - \frac{1}{2}hf(t_i, y_i) + \frac{3}{2}hf(t_{i+1}, y_{i+1})$ .

Given that y(0) = 1 and y(1) = 2 , use the Adams-Bashforth method to approximate y(3) for y = -y^2 with a step size of h = 1.

Possible Answers:

$\frac{133}{8}$

$\frac{97}{8}$

$\frac{33}{4}$

$-\frac{159}{8}$

$\frac{55}{8}$

Correct answer:

$-\frac{159}{8}$

Explanation:

In this problem, we're given two points, so we can start plugging in immediately. If we were not, we could approximate $y_{i+1}$ by using the explicit Euler method on $y_{i}$ .

Plugging into $y_{i+2} = y_{i+1} - \frac{1}{2}hf(t_i, y_i) + \frac{3}{2}hf(t_{i+1}, y_{i+1})$ , we have

$y(2) \approx \mu_2 = 2 -\frac{1}{2}\cdot 1 \cdot -1^2 + \frac{3}{2}\cdot 1 \cdot -2^2 = 2 +\frac{1}{2} - 6 = -\frac{7}{2}$

$y(3)\approx \mu_3 = -\frac{7}{2} + \frac{1}{2}\cdot4 - \frac{3}{2}\cdot \frac{49}{4} =-\frac{159}{8}$ .

Note, our approximation likely won't be very good with such large a time step, but the process doesn't change regardless of the accuracy.

$y(3)\approx -\frac{159}{8}$

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Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Find the solutions to the second order boundary-value problem. y'' -2y' + 2y = 0 , y(0) = 0 , $y\left(\frac{\pi}{2}\right) = 1$ .

Possible Answers:

$y = e^{t}\sin(t) - e^t\cos(t)$

$y = 5e^t\sin(t) + 3e^t\cos(t)$

There are no solutions to the boundary value problem.

$y = 10e^{2t}\sin(2t)$

$y = e^{t -\frac{\pi}{2}}\sin(t)$

Correct answer:

$y = e^{t -\frac{\pi}{2}}\sin(t)$

Explanation:

The characteristic equation of y'' -2y' + 2y = 0 is $\lambda^2 - 2\lambda + 2 = 0$ , with solutions of $\frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$ . Thus, the general solution to the homogeneous problem is $y = c_1e^t\sin(t) + c_2e^t\cos(t)$ . Plugging in our conditions, we find that y(0) = c_2 = 0 , so that $y = c_1e^t\sin(t)$ . Plugging in our second condition, we find that $y\left(\frac{\pi}{2}\right) = c_1e^{\frac{\pi}{2}} = 1$ and that $c_1 = e^{-\frac{\pi}{2}}$ .

Thus, the final solution is $y = e^{t -\frac{\pi}{2}}\sin(t)$ .

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Example Question #1 : Second Order Boundary Value Problems

Find the solutions to the second order boundary-value problem. y'' + 9y = 0 , y(0) = 0 , $y\left(2\pi\right) = 1$ .

Possible Answers:

There are no solutions to the boundary value problem.

$y = 3\sin(3t)$

$y = \sin(3t) + 2 \cos(3t)$

$y = 3\cos(3t)$

$y = 4\sin(3t) - \cos(3t)$

Correct answer:

There are no solutions to the boundary value problem.

Explanation:

The characteristic equation of y'' + 9y = 0 is $\lambda^2 + 9 = 0$ with solutions of $\pm 3i$ . This tells us that the solution to the homogeneous equation is $y = c_1\sin(3t) + c_2\cos(3t)$ . Plugging in our conditions, we find that y(0) = c_2 = 0 so that $y = c_1\sin(3t)$ . Plugging in our second condition, we have $y(2\pi) = 0 = 1$ which is obviously false.

This problem demonstrates the important distinction between initial value problems and boundary value problems: Boundary value problems don't always have solutions. This is one such case, as we can't find c_1, c_2 that satisfy our conditions.

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Example Question #1 : Homogeneous Linear Systems

Find the general solution to the given system.

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

Possible Answers:

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

$X=c_1\binom{1}{3}e^{3t}+c_2\binom{1}{1}e^{-t}$

$X=c_1\binom{-1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

$X=c_1\binom{1}{-1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Correct answer:

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Explanation:

To find the general solution to the given system

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

first find the eigenvalues and eigenvectors.

$det(A-\lambda I )=\begin{vmatrix} 1-\lambda&2 \\ 2&1-\lambda \end{vmatrix}$

$\\=(1-\lambda)(1-\lambda)-(2)(2) \\=1-2\lambda+\lambda^2-4 \\=\lambda^2-2\lambda-3 \\=(\lambda-3)(\lambda+1)$

Therefore the eigenvalues are

$\lambda_1=3, \lambda_2=-1$

Now calculate the eigenvectors

$(A-\lamda I)K=0$

For

$\lambda_1=3$

$\\(1-3)k_1+2k_2 \\2k_1+(1-3)k_2$

Thus,

$k_1=k_2 \rightarrow K_1\binom{1}{1}$

For

$\lambda_1=-1$

$\\(2--1)k_1+1k_2 \\1k_1+(2--1)k_2$

Thus

$k_1=-\frac{1}{3}k_2\rightarrow K_2\binom{-1}{3}$

Therefore,

$X_1=\binom{1}{1}e^{3t};\ X_2\binom{-1}{3}e^{-t}$

Now the general solution is,

$\\X=c_1X_1+c_2X_2 \\\\X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

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Example Question #1 : System Of Linear First Order Differential Equations

Solve the initial value problem $\mathbf{x'} = \begin{bmatrix} 3&-2 \\ -1&2 \end{bmatrix}\mathbf{x}$ . Where $\mathbf{x}_0 = \begin{bmatrix} 1\\-1 \end{bmatrix}$

Possible Answers:

$\mathbf{x} = e^t\begin{bmatrix} 3\\1 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\6 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} -2\\1 \end{bmatrix} +e^{3t}\begin{bmatrix} 2\\2 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} 1\\4 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\2 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} 5\\2 \end{bmatrix} +e^{3t}\begin{bmatrix} -1\\3 \end{bmatrix}$

Correct answer:

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

Explanation:

To solve the homogeneous system, we will need a fundamental matrix. Specifically, it will help to get the matrix exponential. To do this, we will diagonalize the matrix. First, we will find the eigenvalues which we can do by calculating the determinant of $\lambda I - A$ .

$|\lambda I - A| = \begin{vmatrix} \lambda - 3&2 \\ 1&\lambda - 2 \end{vmatrix} = (\lambda -3)(\lambda -2) - 2 = \lambda^2 - 5\lambda +4 = (\lambda -4)(\lambda -1)$

Finding the eigenspaces, for lambda = 1, we have

$\begin{bmatrix} -2&2 &\vline &0 \\ 1&-1 &\vline & 0 \end{bmatrix}$

Adding -1/2 Row 1 to Row 2 and dividing by -1/2, we have $\begin{bmatrix} 1&-1 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$ which means x_1 + x_2 = 0

Thus, we have an eigenvector of $\begin{bmatrix} 1\\1 \end{bmatrix}$ .

For lambda = 4

$\begin{bmatrix} 1&-2 &\vline &0 \\ -1&2 &\vline & 0 \end{bmatrix}$

Adding Row 1 to Row 2, we have

$\begin{bmatrix} 1&-2 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$

So x_1 - 2x_2 = 0 with an eigenvector $\begin{bmatrix} 2\\1 \end{bmatrix}$ .

Thus, we have $P = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}$ and $D = \begin{bmatrix} 1&0 \\ 0&4 \end{bmatrix}$ . Using the inverse formula for 2x2 matrices, we have that $P^{-1} = \begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}$ . As we know that $e^{tA} = Pe^{tD}P^{-1}$ , we have $e^{tA} = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}\begin{bmatrix} e^t&0 \\ 0&e^{4t} \end{bmatrix}\begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}= \begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}$

The solution to a homogenous system of linear equations is simply to multiply the matrix exponential by the intial condition. For other fundamental matrices, the matrix inverse is needed as well.

Thus, our final answer is $\begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}\begin{bmatrix} 1\\-1 \end{bmatrix} = \begin{bmatrix} -3e^t + 4e^{4t}\\-3e^t + 2e^{4t} \end{bmatrix}= e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

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Example Question #1 : Homogeneous Linear Systems

Solve the homogenous equation:

y''-6y'+8y=0

With the initial conditions:

y(0)=2

y'(0)=5

Possible Answers:

$y(t)=\frac{e^{-4t}}{2}+\frac{3e^{-2t}}{2}$

none of these answers

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

$y(t)=C_{1}e^{-4t}+C_{2}e^{-2t}$

$y(t)=C_{1}e^{4t}+C_{2}e^{2t}$

Correct answer:

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

Explanation:

So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

r^2-6r+8=0 Which, using the quadratic formula or factoring gives us roots of $r_{1}=4$ and $r_{2}=2$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}e^{4t}+C_{2}e^{2t}$

We have two initial values, one for y(t) and one for y'(t), both with t=0\

So:

$y(0)=2=C_{1}+C_{2}$

$y'(t)=4C_{1}e^{4t}+2C_{2}e^{2t}$ so: $y'(0)=5=4C_{1}+2C_{2}$

We can solve for $C_{2}$ : $2-C_{1}=C_{2}$ Then plug into the other equation to solve for $C_{1}$

$5=4C_{1}+2(2-C_{1})=4C_{1}+4-2C_{1}=2C_{1}+4$

So, solving, we get: $C_{1}=\frac{1}{2}$ Then $C_2=\frac{3}{2}$

This gives a final answer of:

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

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Example Question #1 : System Of Linear First Order Differential Equations

Solve the second order differential equation:

3y''-3y'-6y=0

Subject to the initial values:

y(0)=4

y'(0)=2

Possible Answers:

none of these answers

$y(t)=3e^{2t}-e^{-t}$

$y(t)=2e^{2t}-2e^{-t}$

$y(t)=2e^{2t}+2e^{-t}$

$y(t)=3e^{2t}+e^{-t}$

Correct answer:

$y(t)=2e^{2t}+2e^{-t}$

Explanation:

So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

3r^2-3r-6=0 Which, using the quadratic formula or factoring gives us roots of $r_{1}=2$ and $r_{2}=-1$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}e^{2t}+C_{2}e^{-t}$

We have two initial values, one for y(t) and one for y'(t), both with t=0

So:

$y(0)=4=C_{1}+C_{2}$

$y'(t)=2C_{1}e^{2t}-C_{2}e^{-t}$ so: $y'(0)=2=2C_{1}-C_{2}$

We can solve $4-C_{1}=C_{2}$ Then plug into the other equation to solve for $C_{1}$

$2=2C_{1}-(4-C_{1})=2C_{1}+C_{1}-4=3C_{1}-4$

So, solving, we get: $C_{1}=2$ Then C_2=2

This gives a final answer of:

$y(t)=2e^{2t}+2e^{-t}$

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Differential Equations : Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Euler Method

Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Example Question #1 : Euler Method

Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Example Question #1 : Second Order Boundary Value Problems

Example Question #1 : Homogeneous Linear Systems

Example Question #1 : System Of Linear First Order Differential Equations

Example Question #1 : Homogeneous Linear Systems

Example Question #1 : System Of Linear First Order Differential Equations

All Differential Equations Resources

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