SOLUTION: Use the "mixed partials" check to see if the following differential equation is exact. If it is exact find a function F(x,y) whose differential, dF(x,y) gives the differential e

Algebra ->  Equations -> SOLUTION: Use the "mixed partials" check to see if the following differential equation is exact. If it is exact find a function F(x,y) whose differential, dF(x,y) gives the differential e      Log On


   



Question 1184041: Use the "mixed partials" check to see if the following differential equation is exact.
If it is exact find a function F(x,y) whose differential, dF(x,y) gives the differential equation. That is, level curves F(x,y)=C are solutions to the differential equation:
dy/dx=−x^4−2y/2x+4y^2
First rewrite as M(x,y)dx+N(x,y)dy=0
where M(x,y)= My answer is x^4-2y but it's wrong.
and N(x,y)= My answer is -2x+4y^2 (wrong)
If the equation is not exact, enter not exact, otherwise enter in F(x,y) as the solution of the differential equation here:
My answer is -(x^5/5)-2yx-(4/3)y^3=C but it's wrong.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Next time, try to make proper use of parentheses. I will assume that your D.E. is
dy%2Fdx+=+%28-x%5E4+-2y%29%2F%282x+%2B+4y%5E2%29.

<===> %28x%5E4+%2B2y%29dx+%2B+%282x+%2B+4y%5E2%29dy=0, with M%28x%2Cy%29+=+x%5E4+%2B2y and N%28x%2Cy%29+=2x+%2B+4y%5E2.
As M%5By%5D+=+2+=+N%5Bx%5D, the D.E. is exact.
As such, you can retrieve the potential function F(x,y) by setting ∂F/∂x = M(x,y) and ∂F/∂y = N(x,y).

In the end what you will get is F%28x%2Cy%29+=+x%5E5%2F5+%2B+2yx+%2B+%284%2F3%29y%5E3+=+C+
Can you see now where you went wrong?