- #1
etotheipi
- Homework Statement
- Solve ##4xy + 1 + (2x^2 + \cos{(y)})y' = 0##
- Relevant Equations
- N/A
I let ##M = 4xy + 1## and ##N = 2x^2 + \cos{(y)}##. Since ##\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}##, the equation is exact and we have $$\frac{\partial f(x,y)}{\partial x} = 4xy + 1$$ From inspection, you can tell this has to lead to $$f(x,y) = 2x^2 y + x + h(y)$$ and we could then derive it partially wrt. ##y## and find out what ##h(y)## is to get the solution. My question is how do we do the integration "formally"; is there such a thing as partial integration? $$\int \partial f(x,y) = \int (4xy + 1) \partial x$$ I say this because if instead we write $$\int d f(x,y) = \int (4xy + 1) dx$$ then we cannot evaluate the RHS since ##y = y(x)##. Thank you!