Doubt regarding a differential equation. Solve: \(\displaystyle{x},{\left.{d}{x}\right.}+{y},{\left.{d}{y}\right.}={\frac{{{a}^{{2}}{\left({x},{\left.{d}{y}\right.}-{y},{\left.{d}{x}\right.}\right)}}}{{{x}^{{2}}+{y}^{{2}}}}}\) The author

ropowiec2gkc

ropowiec2gkc

Answered question

2022-04-02

Doubt regarding a differential equation.
Solve: x,dx+y,dy=a2(x,dyy,dx)x2+y2
The author further proceeds to rearrange above in the form M,dx+N,dy=0 where
M=x+a2yx2+y2;N=ya2xx2+y2
Which further implies
My=Nx=a2(x2y2)(x2+y2)2
This is the required condition for the given equation to be exact and the solution is obtained using standard formula.
But, what I did is as follows:
x,dx+y,dy=a2(x,dyy,dx)x2+y2
(x2+y2)x,dx+(x2+y2)y,dy=a2(x,dyy,dx)
Therefore (x3+xy2+a2y),dx+(y3+yx2a2x),dy=0
Comparing it with the equation M,dx+N,dy=0 we get,
M=x3+xy2+a2y;N=y3+yx2a2x
But My=2xy+a2;Nx=2xya2

Answer & Explanation

Cody Hart

Cody Hart

Beginner2022-04-03Added 11 answers

Step 1
x,dx+y,dy=a2(x,dyy,dx)x2+y2tag1
As you correctly found : Equation (1) is an exact ODE.
You transformed Eq.(1) into Eq.(2):
(x3+xy2+a2y),dx+(y3+yx2a2x),dy=0tag2
Again you correctly found : Equation (2) is not an exact ODE.
Step 2
This is not contradictory because Eqs.(1) and (2) are not the same ODE.
In order to transform Eq.(2) into an exact ODE one have to multiply it with an "integration factor", say μ(x,y).
One find that μ(x,y)=1x2+y2.
μ(x,y)Big((x3+xy2+a2y),dx+(y3+yx2a2x),dyBig)=0tag3
x3+xy2+a2yx2+y2,dx+y3+yx2a2xx2+y2,dy=0is exact. is exact.
In fact Eq.(3) is equivalent to Eq.(1) thanks to the correct μ(x,y).
Aarlenlsi1

Aarlenlsi1

Beginner2022-04-04Added 10 answers

Step 1
In my mind, your question is about integrating factor.
You have checked your equation
(x3+xy2+a2y)dx+(y3+yx2a2x)dy=0 is not exact.
That's why we need integrating factor. Here the integrating factor is exactly 1x2+y2.
Step 2
By the integrating factor, we make your equation an exact equation deliberately, then we can solve the ODE by exact equations.
If x2+y2=0, then this ODE is not well-defined.
p.s. It's my first time to answer a problem. Perhaps I misunderstand your problem, please tell me and I would try to correct my answer.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?