Differential Equations: Find all solutions of these equations

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SUMMARY

This discussion focuses on solving a set of differential equations, specifically: 1) y' + ycos(t) = 0, 2) y' - 2yt = 0, 3) y'tan(t) + ysec2(t) = 0, and 4) y''' = p(x) + 2y where p(x) is an arbitrary polynomial. The participants explore methods such as separation of variables and integration techniques to derive solutions. The conversation highlights the complexity of trigonometric functions in differential equations and emphasizes the need for careful manipulation of terms to isolate variables.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with separation of variables technique
  • Basic knowledge of trigonometric identities
  • Concept of polynomial functions and their derivatives
NEXT STEPS
  • Study the method of separation of variables in depth
  • Learn about integrating factors for first-order linear differential equations
  • Explore polynomial solutions to higher-order differential equations
  • Review trigonometric functions and their derivatives in the context of differential equations
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Students studying differential equations, mathematics educators, and anyone seeking to improve their problem-solving skills in calculus and mathematical analysis.

ballajr
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Homework Statement



1) y' + ycos(t) = 0
2) y' - 2yt = 0
3) y'tan(t) + ysec2(t) = 0

4) Let p(x) be an arbitrary polynomial function. Note that the following is a polynomial
solution to y" = p(x) + y:

y = -p - p" - p(4) - p(6) - ... - p(n)

where n is the smallest positive even integer such that p(n+1) is the zero polynomial.
Prove that the following differential equation has a polynomial solution:

y''' = p(x) + 2y


Homework Equations





The Attempt at a Solution



1) First, I tried moving the ycos(t) to the right, but I don't know where to go from there.
y' = ycos(t)

2) I moved 2ty to the right. Integrated it.
y' = 2ty

3) This one is just a beast. I was never good at trigonometry. I moved the ysec^2t over to the right and canceled out the trig functions to get down to this:
y' = ysec(t)

I don't know where to go from here.

4) I don't even know how to go about doing this problem.
 
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All three of them can be solved by separation of variables. That is if we have a diff. eq of the form:

\frac{dy}{dt}=H(y,t)

we say it is a separable diff. eq. if H(y,t)=\frac{f(t)}{g(y)}.

Try to do simmilar stuff with your particular functions. You've got the first step right for the 1st: moving ycos(t) to the other side(be careful of the sign though). what next? also good job with the 2nd.
 
Oh okay. There's going to be more than one solution for all three of those right? My teacher likes to trick us like that. The second one i haven't solved yet. The fourth one is a bit confusing. I need some help with that.
 

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