Euler Method - Differential Equations

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Question

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

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Answer

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 $\mu(\pi) = \frac{\pi}{2} \approx y(\pi)$ .

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Question

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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Answer

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,

$y(0.2)\approx2.58$

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Question

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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Answer

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,

$y(0.2)\approx2.58$

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Question

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

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Answer

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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Question

Use two steps of Euler's Method with on

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

To three decimal places

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Answer

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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Question

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

Tap to reveal answer

Answer

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 $\mu(\pi) = \frac{\pi}{2} \approx y(\pi)$ .

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Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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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Question

Use two steps of Euler's Method with on

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

To three decimal places

Tap to reveal answer

Answer

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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Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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 $\mu(\pi) = \frac{\pi}{2} \approx y(\pi)$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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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Question

Use two steps of Euler's Method with on

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

To three decimal places

Tap to reveal answer

Answer

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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Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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 $\mu(\pi) = \frac{\pi}{2} \approx y(\pi)$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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,

$y(0.2)\approx2.58$

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Question

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

Tap to reveal answer

Answer

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 $-\frac{1}{16}$

← Didn't Know|Knew It →

Question

Use two steps of Euler's Method with on

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

To three decimal places

Tap to reveal answer

Answer

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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