Solve the differential equation 2y'=yx/(x^2 + 1) - 2x/y

Question

I have to solve the equation: $$ 2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}} $$ I know the first step is to divide by y, which gives the following equation: $$ {\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$

According to my notes I get that I should make a substitution: $$ z = {\frac{1}{y^2}} $$ and the derivative of z:$$ z' = {\frac{y'}{y^3}}$$

But I don't know how to proceed after this... Any help is appreciated!

Answer 1

Hint $$2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}}$$ You can also multiply by y the Bernouilli 's equation $$2y'y= {\frac{xy^2}{x^2+1}} - 2x$$ Observe that $(y^2)'=2y'y$ substitute $z=y^2$ $$z'= {\frac{xz}{x^2+1}} - 2x$$ Now it's a linear first ode

But you can do it your way $${\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$

$${\frac{2y'}{y^2}} ={\frac{x}{y(x^2+1)}} - {\frac{2x}{y^3}}$$

Substitute $z=1/y$ $$-2z' =\frac{zx}{(x^2+1)} - 2xz^3$$

Answer 2

write your equation in the form $$2\frac{dy(x)}{dx}y(x)-\frac{xy(x)^2}{x^2+1}=-2x$$ substituting $$v(x)=y(x)^2$$ and $$\mu(x)=e^{\int\frac{-x}{\sqrt{x^2+1}}dx}=\frac{1}{\sqrt{x^2+1}}$$ we get $$\frac{\frac{dv(x)}{dx}}{\sqrt{x^2+1}}-\frac{xv(x)}{\sqrt{x^2+1}^{3/2}}=\frac{2x}{\sqrt{x^2+1}}$$ and we get $$\frac{\frac{d v(x)}{dx}}{\sqrt{x^2+1}}+\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+1}}\right)v(x)=-\frac{2x}{\sqrt{x^2+1}}$$ and now $$\int\frac{d}{dx}\left(\frac{v(x)}{\sqrt{x^2+1}}\right)dx=\int-\frac{2x}{\sqrt{x^2+1}}dx$$