Non-linear first order ODE with auxiliary variable

Question

I have a problem with this equation: $y'(x)=\frac{2y(x)-x}{2x-y(x)}$.

Using $y=xz$ i'm arrived to prove that $\frac{z-1}{(z+1)^{3}}=e^{2c}x^{2}$, but now i'm stuck.

How can i explain the $z$? I've tried with the substitution but without results...

Thanks for any help!

Answer 1

Writing $$y'(x)=\frac{\frac{2y}{x}-1}{2-\frac{y}{x}}$$ and make the Substitution $$\frac{y}{x}=u$$ and we get $$xu'=\frac{-1-u^2}{2-u}$$ then write $$-\frac{2-u}{1+u^2}du=\frac{1}{x}dx$$

Answer 2

Using $y=xz$ and $y'=z+xz'$, I can write:

$z+xz'=\frac{2xz-x}{2x-xz}=\frac{2z-1}{2-z}\rightarrow xz'=\frac{2z-1}{2-z}-z\rightarrow z'=\frac{z^2-1}{2-z}\cdot \frac{1}{x}\Rightarrow \int \frac{2-z}{z^2-1}=\int dx$

So:

$\frac{2-z}{z^2-1}=\frac{2-z}{(z+1)(z-1)}=\frac{A}{z+1}+\frac{B}{z-1}=\frac{z(A+B)-A+B}{(z+1)(z-1)}\Rightarrow \left\{\begin{matrix} -A+B=2\\ A+B=-1 \end{matrix}\right.\left\{\begin{matrix} A=-\frac{3}{2}\\ B=\frac{1}{2} \end{matrix}\right.$

Therefore:

$\int \frac{2-z}{(z+1)(z-1)}=\int \frac{-\frac{3}{2}}{z+1}dz+\int \frac{\frac{1}{2}}{z-1}dz\rightarrow -\frac{3}{2}log(z+1)+\frac{1}{2}log(z-1)=log(x)+c$

$\frac{1}{2}[log(z-1)-3log(z+1)]=log(x)+c\Rightarrow \frac{1}{2}log(\frac{z-1}{(z+1)^{3}})=log(x)+c$

and I obtain

$\frac{z-1}{(z+1)^{3}}=e^{2(log(x)+c)}\Rightarrow \frac{z-1}{(z+1)^{3}}=e^{2c}x^{2}$

Answer 3

$$\frac {z-1}{(z+1)^3}=Kx^2$$ Substitute back $z=y/x$ and $e^{2c}=K$ $$\frac {(y-x)}{x}\frac {x^3}{(y+x)^3}=Kx^2$$ $$\frac {y-x}{(y+x)^3}x^2=Kx^2$$ $$\frac {y-x}{(y+x)^3}=K$$

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Another method for solving this equation $$y'(x)=\frac{2y(x)-x}{2x-y(x)}$$ $$(2x-y)dy=(2y-x)dx$$ $$\frac {dy}{(2y-x)}=\frac {dx}{(2x-y)}$$ $$\frac {d(y+x)}{(y+x)}=\frac {d(y-x)}{3(y-x)}$$ After integration, $$3\ln (y+x)=\ln (y-x)+K$$ $$\boxed{(y+x)^3=K(y-x)}$$