Is This the Correct Approach to Solving the Exact Differential Equation?

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In summary, the conversation discusses two approaches to solving the differential equation 2(y^2+1)dx+(4xy-3y^2)dy=0. One approach involves observing that it is an exact equation, while the other involves dividing both sides by x^2 and substituting z=y/x. The conversation also addresses a solution that was provided and suggests typing out the solution for easier checking.
  • #1
Baconslider
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Homework Statement


2(y^2+1)dx+(4xy-3y^2)dy=0

Homework Equations

The Attempt at a Solution


here is my attempt to solve
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  • #2
That's one way of solving it (observing that it is exact). I didn't check all steps though, so don't know if you are correct. You can fill in your solution in the differential equation and see if it works.

An alternative approach would be to divide both sides by ##x^2## and substitute ##z= y/x##
 
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  • #3
Math_QED said:
That's one way of solving it (observing that it is exact). I didn't check all steps though, so don't know if you are correct. You can fill in your solution in the differential equation and see if it works.

An alternative approach would be to divide both sides by ##x^2## and substitute ##z= y/x##
im sorry but there is no z in the problem , i think you've mistaken 2 for z sorry about my penmanship :) lol
 
  • #4
Baconslider said:
im sorry but there is no z in the problem , i think you've mistaken 2 for z sorry about my penmanship :) lol

No, you are misunderstanding me.

You can divide both sides by ##x^2## and you will get an expression in ##y/x##. You can consider ##y/x## as a new variable, which we will call ##z##. Thus, ##z = y/x## and ##xz = y##, so ##dy = dx z + xdz## and you can use this to transform your differential equation to one in ##z(x)##. Here, it is unnecessary as the DE is exact, but sometimes the method I just explained can be useful :)
 
  • #5
Baconslider said:

Homework Statement


2(y^2+1)dx+(4xy-3y^2)dy=0

Homework Equations


The result you got is not correct. I can not follow what you did, but M and N are functions of both x and y. When you integrate M(x,y) with respect to x or N(x,y) with respect to y, you get two x,y functions. I do not understand, what you F(x) and F(y) are. Also, why is g'(x)=1, and if it so, why is g(x)=0? You should type in your solution. It would be easier to check what you wrote even for yourself.
 

Related to Is This the Correct Approach to Solving the Exact Differential Equation?

1. How do I know if a problem is an exact differential problem?

An exact differential problem is a type of differential equation where the dependent variable and its derivatives can be written as a total differential of a function. This means that the problem can be solved by finding the function and integrating it.

2. What is the general method for solving an exact differential problem?

The general method for solving an exact differential problem involves identifying the dependent variable and its derivatives, checking for exactness by comparing the mixed partial derivatives, and then finding the integrating factor to solve for the function. Finally, the solution is obtained by integrating the function and adding a constant of integration.

3. How do I find the integrating factor for an exact differential problem?

The integrating factor for an exact differential problem can be found by dividing the coefficient of the differential term in the equation by the coefficient of the dependent variable. This factor is then used to multiply the entire equation, making it exact and allowing for the solution to be found.

4. Can an exact differential problem have multiple solutions?

Yes, an exact differential problem can have multiple solutions. This is because the constant of integration can take on different values, resulting in different functions that satisfy the equation. It is important to always include the constant of integration when solving an exact differential problem.

5. Are there any tips for solving exact differential problems more efficiently?

One tip for solving exact differential problems more efficiently is to always check for exactness before attempting to solve. This can save time and effort, as some problems may not be exact and require a different method of solving. Additionally, practicing with different types of exact differential problems can help improve problem-solving skills and speed.

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