2x dy/dx + 3y + (x+1)y^3 = 0
whado i do????
i try to get it in the form
dy/dx + fxy = gxy^n but it isnt working
im geting loads of denominators which cant be solved
please help with a "hint"
cheers
Have you used the standard substitution $\displaystyle v=y^{-2}$ ?. In such case, what do you obtain?
Fernando Revilla
Yes, it is. Now solve the homogeneous equation $\displaystyle dv/dx-3v/x=0$ or equivalently $\displaystyle dv/v-3dx/x=0$ .
Fernando Revilla
it cannot be homogeneous. this example is not homogenous as it doesnt require knowledge of homogeneous d.es?
yes i'm going to go with the i.f
my log rules need some sharpening up, i didn't realise it came to x^-3, i ended up with just 3 which i knew had to be wrong as it was useless.
how do you integrate -3/x? do you just take the -3 out and integrate 1/x?
i laughed so bad when i saw the product rule thing. this is a really nice question.
the question is what can be done about the RHS because i dont know what to do with it at all
You should have gotten $\displaystyle \displaystyle \frac{d}{dx}(\textrm{something}) = \frac{x+ 1}{x}$, which is equivalent to
$\displaystyle \displaystyle \textrm{something} = \int{\frac{x + 1}{x}\,dx}$.
Also note that $\displaystyle \displaystyle \frac{x + 1}{x} = 1 + \frac{1}{x}$.
And to answer your other question, $\displaystyle \displaystyle \int{-\frac{3}{x}\,dx} = -3\int{\frac{1}{x}\,dx}$, like you suspect.