Numerical Solutions of Ordinary Differential Equations - Differential Equations

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Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

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Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

The formula for Euler approximations $\mu(t+h) = \mu (t) + h \cdot f(t, \mu(t))$ .

Plugging in, we have

$\mu($\frac{\pi}{4}$) = \mu (0) + $\frac{\pi}{4}$f(0, \mu(0)) = 0 + $\frac{\pi}{4}$(\cos(0)(2-0)) = $\frac{\pi}{2}$$

$\mu($\frac{\pi}{2}$) = \mu ($\frac{\pi}{4}$) + $\frac{\pi}{4}$f($\frac{\pi}{4}$, \mu($\frac{\pi}{4}$)) = $\frac{\pi}{2}$ + $\frac{\pi}{4}$(\cos($\frac{\pi}{2}$)(2-$\frac{\pi}{2}$)) = $\frac{\pi}{2}$ + 0 = $\frac{\pi}{2}$$

Here we can see that we've gotten trapped on a horizontal tangent (a failing of Euler's method when using larger time steps). As the function is not dependent on t, we will continue to move in a horizontal line for the rest of our Euler approximations. Thus $\mu(\pi) = $\frac{\pi}{2}$ \approx y(\pi)$ .

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

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Using Euler's Method for the function

first make the substitution of

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

therefore

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.

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Using Euler's Method for the function

first make the substitution of

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

therefore

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

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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 two steps of Euler's Method with on

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

To three decimal places

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Euler's Method gives us

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$

The two-step Adams-Bashforth method of approximation uses the approximation scheme .

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

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

Plugging into , 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}$$

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

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The characteristic equation of is , 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. , , $y\left(2\pi\right) = 1$ .

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The characteristic equation of is 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.

Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

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Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

The formula for Euler approximations $\mu(t+h) = \mu (t) + h \cdot f(t, \mu(t))$ .

Plugging in, we have

$\mu($\frac{\pi}{4}$) = \mu (0) + $\frac{\pi}{4}$f(0, \mu(0)) = 0 + $\frac{\pi}{4}$(\cos(0)(2-0)) = $\frac{\pi}{2}$$

$\mu($\frac{\pi}{2}$) = \mu ($\frac{\pi}{4}$) + $\frac{\pi}{4}$f($\frac{\pi}{4}$, \mu($\frac{\pi}{4}$)) = $\frac{\pi}{2}$ + $\frac{\pi}{4}$(\cos($\frac{\pi}{2}$)(2-$\frac{\pi}{2}$)) = $\frac{\pi}{2}$ + 0 = $\frac{\pi}{2}$$

Here we can see that we've gotten trapped on a horizontal tangent (a failing of Euler's method when using larger time steps). As the function is not dependent on t, we will continue to move in a horizontal line for the rest of our Euler approximations. Thus $\mu(\pi) = $\frac{\pi}{2}$ \approx y(\pi)$ .

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

Tap to see back →

Using Euler's Method for the function

first make the substitution of

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

therefore

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.

Tap to see back →

Using Euler's Method for the function

first make the substitution of

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

therefore

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

Tap to see back →

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 two steps of Euler's Method with on

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

To three decimal places

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Euler's Method gives us

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$

The two-step Adams-Bashforth method of approximation uses the approximation scheme .

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

Tap to see back →

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

Plugging into , 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}$$

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

Tap to see back →

The characteristic equation of is , 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. , , $y\left(2\pi\right) = 1$ .

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The characteristic equation of is 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.

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

Tap to see back →

Using Euler's Method for the function

first make the substitution of

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

therefore

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,

$y(0.2)\approx2.58$

Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

Tap to see back →

Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = $\frac{\pi}{4}$$ and .

The formula for Euler approximations $\mu(t+h) = \mu (t) + h \cdot f(t, \mu(t))$ .

Plugging in, we have

$\mu($\frac{\pi}{4}$) = \mu (0) + $\frac{\pi}{4}$f(0, \mu(0)) = 0 + $\frac{\pi}{4}$(\cos(0)(2-0)) = $\frac{\pi}{2}$$

$\mu($\frac{\pi}{2}$) = \mu ($\frac{\pi}{4}$) + $\frac{\pi}{4}$f($\frac{\pi}{4}$, \mu($\frac{\pi}{4}$)) = $\frac{\pi}{2}$ + $\frac{\pi}{4}$(\cos($\frac{\pi}{2}$)(2-$\frac{\pi}{2}$)) = $\frac{\pi}{2}$ + 0 = $\frac{\pi}{2}$$

Here we can see that we've gotten trapped on a horizontal tangent (a failing of Euler's method when using larger time steps). As the function is not dependent on t, we will continue to move in a horizontal line for the rest of our Euler approximations. Thus $\mu(\pi) = $\frac{\pi}{2}$ \approx y(\pi)$ .

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

Tap to see back →

Using Euler's Method for the function

first make the substitution of

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

therefore

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

Tap to see back →

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}$$