Question 1143624
{{{(x^4 +2y)dx-xdy=0}}}
<pre>
We cannot separate the variables, so we hope that it is linear.
It is linear if we can get it in the form

{{{dy+P(x)y*dx=Q(x)*dx}}}

then we can solve it by multiplying by the
integrating factor {{{matrix(2,1,"",e^int(P(x)*dx))}}}

So let's try to get it in that form:

{{{(x^4 +2y)dx-xdy=0}}}

{{{x^4*dx+2y*dx-xdy=0}}}

{{{-xdy+2y*dx=-x^4*dx}}}

Divide through by -x

{{{-xdy/(-x)+2y*dx/(-x)=(-x^4*dx)/(-x)}}}

{{{dy+(-2/x)y*dx=x^3*dx}}}

So it is a linear differential equation with {{{P(x)=(-2/x)=-2x^(-1)}}} and {{{Q(x)=x^3}}}

Linear differential equations are usually easier if we can
avoid denominators by using negative exponents:

{{{dy+(-2x^(-1))y*dx=x^3*dx}}}

We calculate the integrating factor

{{{matrix(2,1,"",e^int(P(x)^""*dx))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^int((-2x^(-1))*dx))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^(-2*int((1/x)*dx)))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^(-2*ln(x)))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^(ln(x^(-2))))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",x^(-2)))}}}

We multiply through by the integrating factor {{{x^(-2)}}}

{{{x^(-2)*dy+x^(-2)(-2x^(-1))y*dx=x^(-2)*(x^3*dx)}}}

{{{x^(-2)*dy+(-2x^(-3))y*dx=x*dx)}}}

We integrate both sides:

{{{int((x^(-2)*dy+(-2x^(-3))^""y*dx))}}}{{{""=""}}}{{{int(x^""^""*dx))}}}

The right side is easy to integrate.  The left side requires a little
more thinking since we cannot integrate the terms separately.  On the left 
side, we have the integral of the differential of a product 
d(u*v) = u*dv+v*du, where {{{u=x^(-2)}}} and {{{v=y}}}, so the left side
integrates to the product {{{u*v}}} or {{{x^(-2)*y}}} and so the general
solution is

{{{x^(-2)y=x^2/2^""+C}}}

Multiply both sides by x²

{{{x^2*x^(-2)y=x^2*expr(x^2/2^"")+x^2*C}}}

{{{x^0*y=expr(x^4/2^"")+Cx^2}}}

{{{y=x^4/2^""+Cx^2}}}

Edwin</pre>