2x^3ydx+(x^4+y^4)dy" how to solve it in homogeneous method
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1 Answer

I think this is supposed to be:

2x³ydx+(x⁴+y⁴)dy=0, because it would not make sense otherwise.

If we make the substitution y=vx then dy/dx=v+xdv/dx, where v is a function of x.

We then have:

2vx⁴dx+(x⁴+v⁴x⁴)dy=0 and assuming at this point that x≠0, we can divide through by x⁴:


Now replace dy/dx by v+xdv/dx:

v+xdv/dx=-2v/(1+v⁴). We can separate variables x and v:


This can be written:

[(1+v⁴)/(3v+v⁵)]dv=-dx/x ready for integration.

(1+v⁴)/(3v+v⁵) is (1+v⁴)/(v(3+v⁴)) and split into partial fractions:

A/v+Bv³/(3+v⁴) where A and B are constants to be found:

A(3+v⁴)+Bv⁴≡1+v⁴, so equating coefficients: A+B=1 (v⁴), 3A=1 (constant), so A=⅓ and B=⅔.

We need to solve:


The left hand integral can be split:


If we let u=3+v⁴, du=4v³dv, so v³dv=du/4.

So we have:

⅓ln|v|+⅔∫(1/u)(du/4)=-ln|x|+c where c is constant of integration.

⅓ln|v|+⅙ln|u|=ln|a/x| where c=ln(a), that is, a=e^c, a constant.


ln(v²(3+v⁴))=ln(A/x⁶), where A=a⁶,


But v=y/x, so (y/x)²(3+(y/x)⁴)=A/x⁶,



by Top Rated User (721k points)

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