Solution of differential Equation.

Question

Here's the function that is needed to be solved,

$$ y' = 2 \ - \frac{y}{x}$$

I tried following, Substituting $\frac{y}{x}= t$ so that $ y' = t +x\cdot t' $

So, $$ t+ x\cdot t' = 2\ -t $$

$$2t +x\cdot t' =2 $$

Where $y'= \frac{dy}{dx} \ and \ t' = \ \frac{dt}{dx}$

Now I don't now how to proceed further. Any help is appreciated. Thanks.

Answer 1

Substituting $$y=ux$$ we will get $$y'=u'x+u$$ so $$u'x=2-2u$$ $$\frac{du}{2-2u}=\frac{dx}{x}$$

Answer 2

I'll mention another approach. The equation is of the form $y'+Py=Q$ with $P=x^{-1},\,Q=2$. If $R:=\exp\int Pdx$ then $R'=RP$ and $(Ry)'=R(y'+Py)=RQ$, so $y=R^{-1}\int RQ dx$. (Without loss of generality $R$ may be multiplied by a constant, but that won't change the solution set.) In this case we may take $R=x$, so $y=x^{-1}\int 2xdx=x+Cx^{-1}$.

Answer 3

you are on the correct path

The equation is separable

$$2t +x\cdot t' =2 \implies xt'=2-2t \implies \int \frac {dt}{1-t}=2\int \frac {dx}{x}=2\ln|x|+K$$ $$......$$

Another approach $$y' = 2 \ - \frac{y}{x} \implies y'+\frac yx=2$$ Multiply by x $$y'x+y=2x \implies (xy)'=2x$$ Integrate $$xy=2\int x dx=x^2+K$$ $$y(x)=x+\frac Kx$$