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My Work:

IMG_3410.jpg

My Questions:

4a) How do you go from dy = -u^-2du to dy/dx = -u^-2dy/dx?

4b) Why are you allowed to remove the absolute values from u+1?

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    For part (a), if $u = 1/y$, then $y = 1/u$. Substitute $1/u$ for $y$ in the original equation and simplify (using the rules of derivatives, including the chain rule). – 2017-02-22
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    Also, is this a take-home exam? – 2017-02-22
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    No this is a practice prelim that I cannot figure out – 2017-02-22
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    Well, at least do part (a), and show us your work. – 2017-02-22
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    Please attach a comment when you downvote – 2017-02-22
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    I have not downvoted you but I think if you do what quasi said, it would be an improvement to your question. – 2017-02-22
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    I didn't downvote you but ... (1) don't use all CAPS in your complaint. (2) It's a HW type problem and you didn't show _any_ work at all. – 2017-02-22
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    After you have part 1) done, part 2) is can be done simply as it's a separable differential equation. – 2017-02-22
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    How about you include more context in your post (via your own fingers on your own keyboard, and not via an attachment) before you ask "why am I getting downvoted"? – 2017-02-22
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    @amWhy I saw the not-so-nice comment you posted previously before deleting it. Please don't say something like that again. I will post what I have. I had worked on this question previously for an hour, but was too lazy to put up context to this question. I see the consequences of doing so and next time, I will. – 2017-02-22
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    Let me suggest you review your three posts to MSE in the last 18 hours. Interestingly, in total, you've asked for help with 70%, in total of the assignment, test, practice test, while showing no work or attempts on your part. – 2017-02-22
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    @amWhy They are all practice prelims. My university does not do take home exams for math. That is genuinely because I did not know how to solve them after trying them for an hour. Those people's answer's truly helped me understand the problem when I didn't have office hours to go to, so please don't downvote those questions. I can attach a picture of my work to those questions, but they don't make any sense. – 2017-02-22
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    I will not nor have I downvoted your questions. I hope you've gained some understanding about the concepts involved in your questions. It's just that we can all help you so much more when we know what you tried, and where you got stuck. The most important thing ultimately, is facilitating your understanding so you feel confident enough to go forward. – 2017-02-22
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    okay I understand, I will do that going forward thank you! – 2017-02-22

1 Answers 1

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For the first part, note that $u=\frac{1}{y} \Rightarrow y=\frac{1}{u}$. Therefore differentiating gives $dy=-\frac{1}{u^2}~du$.

Substitute this into your ODE: $$\frac{dy}{dx}+2xy=-2xy^2$$ $$-\frac{1}{u^2}\cdot \frac{du}{dx}+2x\frac{1}{u}=-2x\frac{1}{u^2}$$ You can now multiply both sides by $-u^2$ to obtain the desired linear ODE.

As pointed out by @Nadiels, you can solve part (b) as a separable differential equation. You can factor your original ODE as:

$$\frac{du}{dx}=2x(u+1)$$

I'll leave you to separate it and solve for $u(x)$.

For part (c), just use the solution for $u(x)$ you have on part (b) and substitute $u=\frac{1}{y}$.

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    becomes separable doesn't it? – 2017-02-22
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    @Nadiels Good point. Thank you. I'll add this to my answer. – 2017-02-22
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    @Nadiels I used your advice and I still have a few questions about part a and b. For (a), how do you go from dy = -u^-2du to dy/dx = -u^-2dy/dx? For (b), why are you allowed to remove the absolute values from u+1? – 2017-02-22
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    For the first question, I did: $$\frac{dy}{dx}=\frac{\left(-\frac{1}{u^2}~du\right)}{dx}=-\frac{1}{u^2}\cdot \frac{du}{dx}$$ For the second question, see [this](http://math.stackexchange.com/questions/311677/question-regarding-usage-of-absolute-value-within-natural-log-in-solution-of-dif). – 2017-02-23
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    By the way, the solution you've posted is correct. – 2017-02-23
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    @projectilemotion epic thanks! – 2017-02-23