Show the Exact Differential Equation solution is also a solution to another equation

In summary: Is it possible that you're using a browser that doesn't support the x2 tag?Anyway, it looks like you're all set. :)So in summary, the conversation discusses an equation and its exactness, as well as finding a solution for the original equation using a solution from a new exact equation. Part d is also mentioned, regarding whether any solutions were lost in the process.
  • #1
sunnyceej
15
0

Homework Statement


Consider the equation (y^2+2xy)dx-x^2dy=0 (a) Show that this equation is not exact. b) Show that multiplying both sides of the equation by y^-2 yields a new equation that is exact. C) use the solution of the resulting exact equation to solve the origional equation. d) Were any solutions lost in the process?

Homework Equations


How do you show that the solution in part b is a solution to the original?


The Attempt at a Solution


answer to part b: (x+x^2y^-1=c)
I tried (y^2+2xy)dx - x^2dy=(x+x^2y^-1) and integrated both sides to get
xy^2 + x^2y - x^2y = xy+x^2 ln y

I tried solving (x+x^2y^-1=c) for x: getting x=-x^2y^-1 +c and plugging in into the original equation : (y^2+2(-x^2y^-1 +c)y)dx - (-x^2y^-1 +c)^2dy=0 and that just got me a mess. I'm not sure what else to try.
I'm not sure how to do part d, either, but I figured I needed part c first.
 
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  • #2
welcome to pf!

hi sunnyceej! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
sunnyceej said:
answer to part b: (x+x^2y^-1=c)

yes :smile:

but after that, i don't understand what you're doing :confused:

just rewrite x + x2/y = C in the form y = … :wink:
 
  • #3


Thanks! I got y=x^2/(c-x)

Every time I use the x2 button at the top I get SUP/SUP

I thought I also had to plug it back into the original to verify, but I found out solving for y is good! :)
 
Last edited:
  • #4


sunnyceej said:
Thanks! I got y=x^2/(c-x)

Every time I use the x2 button at the top I get SUP/SUP

As a result, your x[ SUP]2[ /SUP] looks like x2 as it should.
 

1. How can a solution to one differential equation also be a solution to another?

The concept of an exact differential equation states that the solution to the equation is independent of the path taken to reach that solution. This means that if two differential equations have the same solution, then the solution to one equation will also be a solution to the other.

2. Are there any restrictions on the types of equations that an exact differential equation solution can be applied to?

Yes, there are certain conditions that must be met for an exact differential equation solution to be applicable to another equation. The equations must have the same form and must have the same independent variables.

3. Can an exact differential equation solution be used to solve any type of equation?

No, exact differential equation solutions are only applicable to equations that can be written in the form of an exact differential equation. This means that the equation must be able to be expressed as the total derivative of a function, with respect to one or more independent variables.

4. How can I determine if a solution to an exact differential equation can also be applied to another equation?

The best way to determine if a solution to an exact differential equation can be applied to another equation is to compare the forms of the two equations. If they have the same form and independent variables, then the solution should be applicable. You can also check if the total derivatives of the two equations are equal.

5. Is there any benefit to using an exact differential equation solution to solve other equations?

Yes, using an exact differential equation solution can save time and effort in solving other equations. It allows for a more efficient and accurate solution, as it eliminates the need for any additional integration or manipulation of the equation. It also provides a deeper understanding of the relationship between different equations and their solutions.

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