Question

Show that the equation is an exact equation and find the general solution

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

Answer 1

take the partial derivatives: Mx=9x^2y^2+4y^3 and Ny=9x^2y^2 +4y^3

Answer 2

anything with a dx take the partial derivative with respect to y, anything with the dy take the partial with respect to x, notice how they are the same, hence exact equations

Answer 3

\[ \psi(x,y)=\int\limits_{}^{}Mdx=\int\limits_{}^{} 3x^{2}y^{3} +y^{4} dy=x^{3}y^{3}+xy^{4}+h(y)\]

Answer 4

is the h(y) the constant?

Answer 5

\[\Psi y=3x^{3}y^{2} +4xy^{3} +h'(y) \]

Answer 6

yes the constant is a function of y because i integrated with respect to dx, its opposites. Now set \[\psi y=N .............where.......3x^{3}y^{2}+4xy^{3}+h'(y)=3x^{3}y^{2}+y^{4}+4xy^{3}\]

Answer 7

h'(y)=y^4 and h(y)=1/5y^5 + f(x)

Answer 8

\[\Psi (x,y) =x^{3}y^{3}+xy^{4} +\frac{1}{5}y^{5}+f(x)\]

Answer 9

where did the y^4 come from on the right hand side of the equation 3 steps up

Answer 10

from 3 steps ago set N=\[\psi y\]

Answer 11

oh ok i get it.

Answer 12

Thanks for the help, i'll go try a few on my own now.