Separation of variables skill builders #178
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mzbush
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This skill builder has done a pretty good job translating the notes. There are three main changes that need to happen before the pull request can be merged.
- All of these changes should be moved down to the bottom of the page after the examples.
- The problems should use the
{solution}directive so that the viewer can click to see the solution. I would recommend looking at the skill builder problems incalculus/series-expansion.mdto get a better idea of how to use the{solution}directive. - There should be some words to help explain the process of solving these equations.
I've added some suggested changes in problem a to improve the presentation of the problem and make it more in line with how the skill builder problems are expected to look. These sorts of changes should be applied to b and c as well.
first-order-odes/separable.md
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| Note that here, we are being careful to denote the redefinition of the | ||
| integration constant ($c = e^{c_0}$). This detail may be glossed over at times. | ||
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| Solve the following: |
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| Solve the following: | |
| ## Skill builder problems | |
| Solve the following: |
This section needs a heading
first-order-odes/separable.md
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| (a) | ||
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| $$ |
$$ is used to wrap equations that are on the same line as normal text. It is not needed here.
first-order-odes/separable.md
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| \begin{alignat*}{3} | ||
| y'+(x+2)y^2 = 0 ; \quad y(1) = 1 \\ | ||
| \frac{dy}{dx} = -(x+2)y^2 \\ | ||
| \int\frac{dy}{y^2} = \int-(x+2)dx \\ | ||
| -\frac{1}{y} = -\left(\frac{x^2}{2} + 2x\right) + C \\ | ||
| y(1) = -1 = -\left(\frac{1}{2} + 2 \right) + C & \quad \to & \quad C = \frac{3}{2} \\ | ||
| \to \quad y = \frac{2}{x^2+4x-3} | ||
| \end{alignat*} | ||
| $$ |
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| \begin{alignat*}{3} | |
| y'+(x+2)y^2 = 0 ; \quad y(1) = 1 \\ | |
| \frac{dy}{dx} = -(x+2)y^2 \\ | |
| \int\frac{dy}{y^2} = \int-(x+2)dx \\ | |
| -\frac{1}{y} = -\left(\frac{x^2}{2} + 2x\right) + C \\ | |
| y(1) = -1 = -\left(\frac{1}{2} + 2 \right) + C & \quad \to & \quad C = \frac{3}{2} \\ | |
| \to \quad y = \frac{2}{x^2+4x-3} | |
| \end{alignat*} | |
| $$ | |
| ```{solution} | |
| \begin{align} | |
| \frac{dy}{dx} &= -(x+2)y^2 \\ | |
| \int\frac{dy}{y^2} &= \int-(x+2)dx \\ | |
| -\frac{1}{y} &= -\left(\frac{x^2}{2} + 2x\right) + C \\ | |
| \end{align} | |
| \begin{equation} | |
| y(1) = -1 = -\left(\frac{1}{2} + 2 \right) + C \quad \to \quad C = \frac{3}{2} | |
| \end{equation} | |
| \begin{equation} | |
| \to \quad y = \frac{2}{x^2+4x-3} | |
| \end{equation} | |
| ``` |
Solving the problem should all go inside a {solution} directive like other skill builder problems. This solution could also use some words to explain what is happening and what each group of expressions is accomplishing in order to get to the solution. In this instance, I have separated the integration, solving for the constant C, and the final answer into groups. There should be some text between these different groups.
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@kaylee0307 please make sure to finalize this and request a review from me ASAP to get credit for it! |
mzbush
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mzbush
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This pull request looks really so far. I have some suggestions to fix up a few minor things, and then once all of them are addressed I think it will be ready for merging.
first-order-odes/separable.md
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| \begin{equation} | ||
| y = \sqrt{ln(x^2-2x+e} | ||
| \end{equation} |
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Should there be a ± here so it's consistent with problem b? Then after stating to choose the positive, the final equation would be restated without the ±.
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Resolves #93
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