Question

Determine whether the given differential equation is Exact or not exact. If it is exact, solve it directly. If it not exact, solve it by finding the appropriate integrating factor.

Answer 1

Answer:

lets take a differential equation

$xy^3dx+(x^2y^2+1)dy=0\\let\\M= xy^3\\and\\N=(x^2y^2+1)$

taking derivative of M w.r.t y and N w.r.t x

$dM/dy=3xy^2\\dN/dx=2xy^2\\so\\dM/dy\neq dN/dx$

so this is an in exact equation we will find integrating factor and make it exact equation

$U=e^\int\limits^_{(-dN/dx-dM/dy)} \,/ M\\\\=e^\int\limits^ {2xy^2-3xy^2} \, /xy^3\\=e^\int\limits^ {-xy^2} /xy^3\\=e^\int\limits^{-1/y} \\=e^-^l^n^y\\=1/y$

so 1/y is our integrating factor multiply it by the equation we will get an exact equation:

$1/y(xy^3)dx+1/y(x^2y^2+1)dy=0\\xy^2dx+(x^2y+1/y)dy=0\\let \\M=xy^2\\and \\N=(x^2y+1/y)$

take derivative

$dM/dy=2xy\\dN/dx=2xy\\so\\dM/dy=dN/dx$

now solving this exact equation

$f(x,y)=c=\int\limits{M} \, dx +g(y)\\=\int\limits {xy^2} \, dx +g(y)\\=\frac{1}{2}x^2y^2+g(y)\\ df/dy=d/dy(\frac{1}{2} x^2y^2+g(y))=N$

$x^2y+g'(y)=x^2y+1/y\\g'(y)=1/y\\g(y)=lny\\f(x,y)=\frac{1}{2}x^2y^2+lny=C$