Numerical Solutions of Ordinary Differential Equations

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Differential Equations › Numerical Solutions of Ordinary Differential Equations

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

There are no solutions to the boundary value problem.

$y = 3\sin(3t)$

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

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

$y = 3\cos(3t)$

Explanation

The characteristic equation of 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 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 that satisfy our conditions.

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

There are no solutions to the boundary value problem.

$y = 3\sin(3t)$

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

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

$y = 3\cos(3t)$

Explanation

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

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

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

There are no solutions to the boundary value problem.

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

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

Explanation

The characteristic equation of 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 , 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 $y = e^{t -\frac{\pi}{2}}\sin(t)$ .

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

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

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

There are no solutions to the boundary value problem.

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

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

Explanation

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

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 and , use the Adams-Bashforth method to approximate for with a step size of

$-\frac{159}{8}$

$\frac{97}{8}$

$\frac{55}{8}$

$\frac{33}{4}$

$\frac{133}{8}$

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.

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

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 and , use the Adams-Bashforth method to approximate for with a step size of

$-\frac{159}{8}$

$\frac{97}{8}$

$\frac{55}{8}$

$\frac{33}{4}$

$\frac{133}{8}$

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.

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

Use the implicit Euler method to approximate for , given that , using a time step of

$-\frac{1}{16}$

$\frac{1}{16}$

$\frac{1}{32}$

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 $-\frac{1}{16}$

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

$y(0.2)\approx2.58$

$y(0.2)\approx2.542$

$y(0.2)\approx2.5$

$y(0.2)\approx2.584$

$y(0.2)\approx2.458$

Explanation

Using Euler's Method for the function

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 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,

$y(0.2)\approx2.58$

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

$y(0.2)\approx2.58$

$y(0.2)\approx2.542$

$y(0.2)\approx2.5$

$y(0.2)\approx2.584$

$y(0.2)\approx2.458$

Explanation

Using Euler's Method for the function

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 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,

$y(0.2)\approx2.58$

Use two steps of Euler's Method with on