Solve a simple differential equation

Question

This is an equation I met while solving a probability theory problem:

$ dy = 2(m/x-y) \cdot dx/x $

m is a constant

Also it is known that $y(m) =1$.

I have the function which satisfies the equation and I verified that it works. But I haven't solved a single differential equation in many years so... I just got curious how one can derive this solution.

Is this equation of some particular type which is solvable through a well-known procedure? Any reference?

Sorry... I am typing on my phone... I will try to improve the equation outlook now.

Answer 1

We have $$\\ \\ \frac { dy }{ dx } =\frac { 2\left( \frac { m }{ x } -y \right) }{ x } =\frac { 2m-2xy }{ { x }^{ 2 } } \\ { y'x }^{ 2 }+2xy=2m\\ d\left( { yx }^{ 2 } \right) =2m\\ y{ x }^{ 2 }=2mx+C\\ y\left( x \right) =\frac { 2mx+C }{ { x }^{ 2 } } \\ \\ \\ \\ $$ and the fact $y\left( m \right) =1$ gives us

$$y\left( m \right) =\frac { 2{ m }^{ 2 }+C }{ { m }^{ 2 } } =1\Rightarrow C=-{ m }^{ 2 }$$

$$y\left( x \right) =\frac { 2mx-{ m }^{ 2 } }{ { x }^{ 2 } } $$

Answer 2

$$\dfrac{dy}{dx}=\dfrac{2m}{x(x-y)}\\\dfrac{dx}{dy}=\dfrac{x^2-xy}{2m}\\ x'+\frac{y}{2m}x=\frac1{2m}x^2$$ It's Bernoulli form, make $z=x^{-1}$ then $\frac{dz}{dy}=-x^{-2}\frac{dx}{dy}$ isolate x' and substitute in the ode, it reduces into a Linear Form do integrating factor and some algebra and you get it in the form $x(y)= F(y) +C$

Answer 3

$$dy = 2(m/x-y) \cdot dx/x$$ $$xdy -2(\frac mx-y)dx=0$$ $$xdxy -(2m-xy)dx=0$$ I's separable $$\int \frac {dxy}{(xy-2m)}=-\int \frac {dx}x$$ $$\ln(xy-2m)=-\ln(x)+K$$ $$(xy-2m)=\frac Kx$$ $$\implies y(x)=\frac {K+2mx}{x^2}$$

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As @haqnatural has pointed out $$xdy -2(m/x-y) dx=0$$ the differential is exact with integrating factor $\mu(x)=x$ $$x^2dy -2(m-xy) dx=0$$ $$\implies \partial_x(x^2)=\partial_y(2xy-2m)$$