![\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5Cleft%282xy%5E4%2B%5Cfrac1%7Bx%2By%5E2%7D%5Cright%29%7D%7B%5Cpartial%20y%7D%3D8xy%5E3-%5Cdfrac%7B2y%7D%7B%28x%2By%5E2%29%5E2%7D)
![\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5Cleft%284x%5E2y%5E3%2B%5Cfrac%7B2y%7D%7Bx%2By%5E2%7D%5Cright%29%7D%7B%5Cpartial%20x%7D%3D8xy%5E3-%5Cdfrac%7B2y%7D%7B%28x%2By%5E2%29%5E2%7D)
so the ODE is indeed exact and there is a solution of the form
. We have
![\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%3D2xy%5E4%2B%5Cdfrac1%7Bx%2By%5E2%7D%5Cimplies%20F%28x%2Cy%29%3Dx%5E2y%5E4%2B%5Cln%28x%2By%5E2%29%2Bf%28y%29)
![\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%3D4x%5E2y%5E3%2B%5Cdfrac%7B2y%7D%7Bx%2By%5E2%7D%3D4x%5E2y%5E3%2B%5Cdfrac%7B2y%7D%7Bx%2By%5E2%7D%2Bf%27%28y%29)
![f'(y)=0\implies f(y)=C](https://tex.z-dn.net/?f=f%27%28y%29%3D0%5Cimplies%20f%28y%29%3DC)
![\implies F(x,y)=x^2y^3+\ln(x+y^2)=C](https://tex.z-dn.net/?f=%5Cimplies%20F%28x%2Cy%29%3Dx%5E2y%5E3%2B%5Cln%28x%2By%5E2%29%3DC)
With
, we have
![8+\ln9=C](https://tex.z-dn.net/?f=8%2B%5Cln9%3DC)
so
![\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2y%5E3%2B%5Cln%28x%2By%5E2%29%3D8%2B%5Cln9%7D)
Answer:
ok so basically...idk how to explain it but i'll give u an example l-8l = 8 or l8l or l-6l = l6l, i don't know if u understand me, but basically it's like the.....opposite of that number or the absolute value
Answer:I want to go to school
Step-by-step explanation:because it is way eaiser