Solving Initial Value Problems for Differential Equations - Tips & Examples

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Homework Help Overview

The discussion revolves around solving initial value problems for differential equations, specifically focusing on techniques such as separation of variables and substitution methods. Participants are exploring their understanding of these concepts and seeking guidance on effective approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve an initial value problem using separation of variables but expresses confusion about the process and the solution provided in the textbook. Another participant confirms the correctness of both the original poster's solution and the book's answer, suggesting a trigonometric identity for clarification. Additionally, a different participant discusses the substitution method and questions the usefulness of identifying degrees of M(x,y) and N(x,y) in their problem-solving process.

Discussion Status

Participants are actively engaging with the problems, offering insights and suggestions for techniques. Some guidance has been provided regarding the use of trigonometric identities and substitution methods, but there is no explicit consensus on the best approach for all scenarios. The discussion reflects a mix of interpretations and attempts to clarify the processes involved.

Contextual Notes

Participants mention a lack of examples in their textbook regarding initial value problems, which may contribute to their uncertainty. There is also a consideration of how to choose between different substitution variables based on the desired outcome.

Gogeta007
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Homework Statement



Im having trouble with "initial value" problems. I can't seem to grasp the "way" around it. any help is greatly appreciated.

\sqrt{}1-y^2 dx - \sqrt{}1-x^2 dy = 0
y(0) = sqr(3) / 2

Homework Equations



solve by separation of variables and inital value.

The Attempt at a Solution



rearranging
i get dy / \sqrt{}1-y^2 = dx / \sqrt{}1-x^2
integrating
arcsin(y) = arcsin(x) + c

y = sin (arcsin (x) + c))
y (o) = sqr3/2

c = pi/3

therefore y = sin(arcsin x + pi/3)

the back of the book says:

y=.5x + sqr(3)/2 (\sqrt{}1-x^2

====
i was reading the book but there's only 1 example about initial value problems.
so If anyone can give me "tips" on how to solve this problems that'd be great
 
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Your answer is correct, but so is the book's. Use sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to show this.
 
Thank you.

Another question regarding my homework, same topic (differential equations)

Im doing "substitution" where you have to identify that M(x,y) and N(x,y) have the same degree by doing M(tx,ty) and N(tx,ty) and such. . . substitute by u= x/y or u=y/x

anyways I don't see any "usefulness" in this process. . .1. Problem:
(x-y)dx + xdy = 0

M = x - y
N = x

u=x/y
x=uy
dx = ydu + udy

subsitute:

(uy - y)(ydu + udy) + uydy = 0 (now everything is more complicated. . .yay?)
multiply trough?
or what do I do next?
y^2 udu + u^2 ydy - y^2 du = 0

and now what?

thnkz again
 
Divide the equation through by xdx. (1-y/x)+dy/dx=0. Set u=y/x and figure out what dy/dx is in terms of du/dx. Eliminate the y's. Not the x's.
 
are there any specific "tips" on like how to work around them? from what i saw there its good to get a dy/dx somewhere (like there by dividing everything by xdx) and then doing the substitution. . .
but how do you know if its better to use u=y/x or v=x/y ?
 
Depends what you want the dependent variable to be. If you want to find y as a function of x, then I suggest u=y/x. If you want x as a function of y then x/y is probably better. Really, just practicing with different combinations will tell you what's better. That's how I learned.
 

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