I have a week trying to solve this DE and I don't find a Integral Factor for it integrate, what I can do to solve?
$\displaystyle 6y^2dx= x(2x^3+y)dy$
I tried reduce to exact DE but I don't find the integral factor, help me
I have a week trying to solve this DE and I don't find a Integral Factor for it integrate, what I can do to solve?
$\displaystyle 6y^2dx= x(2x^3+y)dy$
I tried reduce to exact DE but I don't find the integral factor, help me
Since this involves only powers of x and y I would look for an integrating factor of the form $\displaystyle x^ny^m$. Multiplying by that we have $\displaystyle Mdy+ Ndx= (2x^{m+4}y^n+ x^{m+1}y^{n+1})dy- 6x^my^{n+2}dx= 0$. Then $\displaystyle M_x= 2(m+ 4)x^{m+3}y^n+ (m+1)x^my^{n+1}$ and $\displaystyle N_y= -6(n+2)x^my^{n+1}$. This equation will be exact if and only if those are equal:$\displaystyle M_x= 2(m+ 4)x^{m+3}y^n+ (m+1)x^my^{n+1}= -6(n+2)x^my^{n+1}$. Comparing powers of x, we see that we have $\displaystyle x^my^{n+1}$ on both sides except for that first $\displaystyle x^{m+3}y^n$. We can get rid of that by making the coefficient, 2(m+4), 0: we need m= -4. With m= -4, we must have $\displaystyle -3x^{-4}y^{n+1}= -6(n+2)x^{-4}y^{n+1}$ so we need $\displaystyle -3= -6(n+2)$ so $\displaystyle n+ 2= \frac{1}{2}$ or $\displaystyle n= -\frac{3}{2}$. An integrating factor is $\displaystyle x^{-4}y^{-3/2}$.
\begin{align}
6y^2 \,\mathbb dx &= x(2x^3+y) \, \mathbb dy \\
6\left(\tfrac{y}{x}\right)^2 &= 2x^2 + \tfrac{y}{x}y' \\[8pt]
y &= vx \\
y' &= xv' + v \\[8pt]
6v^2 &= 2x^2 + v(xv'+v) \\
5v^2 &= 2x^2 + xvv' \\
v' - \tfrac5x v &= -2x\tfrac1v
\end{align}
This is a Bernoulli Equation so we can set $w=v^2$ getting $w' = 2vv'$ and thus
\begin{align}
2vv' - \tfrac{10}x v^2 &= -4x \\
w' - \tfrac{10}{x}w &= -4x
\end{align}
Which is a linear equation. (Of course we can get here in one substitution $y^2=wx^2$).
\begin{align}
u(x) &= e^{\int -\frac{10}x\,\mathrm dx} \\ &= e^{-10\ln x} \\ &= x^{-10} \\[8pt]
\tfrac{\mathrm d}{\mathrm dx}\left(x^{-10}w\right) &= -4x^{-9} \\ x^{-10}w &= \tfrac12 x^{-8} + c_1 \\ w &= \tfrac12 x^2 + c_1 x^{10} \\ y^2 &= \tfrac12 x^4 + c_1 x^{12}
\end{align}