[college math: differential equations] what strategy can I use?

[u/The_real_fujiwara]

Stuck on this last problem, hoping someone can give me some advice how to tackle it. I thought it looked like a good equation for exactness but that didn’t work. I asked chat gpt and it said something about shifting the variables so y=y+1 and x=x-1 or something like that but that doesn’t look like anything we’ve covered in class yet. We have done homogeneous equations where I have to substitute in u=y/x and stuff like that but I’m struggling to see how that would help here.

[u/AutoModerator]

Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^{OP and Valued/Notable Contributors can close this post by using /lock command}

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Maximum_Bathroom3490]

The first technique I learned for an equation of the form (ax+by+c)dx + (dx+ey+f)dy = 0 is to find the point of intersection (m,n) of the linear functions (ax+by+c) and (dx+ey+f), let X=x-m and Y=y-n, then find dx and dy as a function of dX and dY
Substituting back, it will yield an homogeneous equation of known form. The proof is quite lengthy but elementary. If the two linear functions are parallel, then another substitution will yield a separable form.

[u/The_real_fujiwara → u/Maximum_Bathroom3490]

Ok thank you, I think this is what chat gpt was trying to get at but it wasn’t making sense to me where the shift was coming from or how I should know what m and n should be besides divine inspiration.

[u/Southlander24]

GPT and the other commenter Maximum_Bathroom are both correct. It becomes homogeneous only after you shift the variables.

Imagine if you had (-4x + y) dy = (2x - 5y) dx instead, without the constants. Then that would be the same as (-4 + y/x) dy/dx = (2 - 5y/x), or (-4 + v) (v + x dv/dx) = (2 - 5v) with the substitution y = vx.

So you are looking for a linear shift where the constant vanishes under the new variables X, Y. If you let X + m = x and Y + n = y, that means -4m + n = -9 and 2m - 5n = 9. Solve this to get m = 2 and n = -1, so X = x - 2 and Y = y - (-1) = y + 1.

I would definitely contact your lecturer about what is examinable in the course. Given that this week's topic was probably homogeneous equations, maybe your lecturer just ran out of time and forgot to teach you about this strategy. But there are a *lot* of special manipulations one can apply to a specific class of problems such that the transformed equation becomes homogeneous.

[u/The_real_fujiwara → u/Southlander24]

Ok thank you I will try this, it makes much more sense where the shift comes from now. The course is an advanced mathematics course for graduate level engineering and this week was just speeding through DE review, we went through everything from separable equations to Bernoulli’s and I wouldn’t be surprised if my professor just skipped a few things since the section on homogeneous equations was like 10 minutes long. This is the first pure math course I’ve had in like 3 years so I’m brushing the rust off.

[u/Southlander24 → u/The_real_fujiwara]

Okay wow, I didn't expect this to be at graduate level, but that makes sense as especially in Asian or European countries this would have been taught in undergraduate. From personal experience, I didn't learn this but I found these strategies okay after doing problem sets.

(There's a Taiwanese maths teacher on a certain adult site which covers this material in Chinese, if you happen to be able to understand the language.)

[u/The_real_fujiwara → u/Southlander24]

This probably was taught at undergrad level just early on and I forgot it, all my courses for the last few years have been engineering focused, where the motto is “let the mathematicians make the formulas and we’ll just use them”

[u/Independent-Mark-162]

Use integrating factor(you can search up the method). I don't remember it properly because I don't like differential equations but when the equation is not exact we use integrating factor to make it exact, you may even be able to derive it, it's pretty straightforward. If I remember correctly you multiply a function(integrating factor that we will know later what it is exactly) then make the differential equation exact then you can do few manipulation to find out what was the thing that made the equation exact.