First order ODE question

Question

$y' = \dfrac{1}{x+2y}$

This might just be a very easy ODE, but i'm getting stuck somehow.

Let $y=v(x)\cdot x \iff y' = v(x) + x\dfrac{dv}{dx}$

And then $\Rightarrow v(x) +x \dfrac{dv}{dx} = \dfrac{1}{x+2x\cdot v(x)} \iff \dfrac{dv}{dx} = \dfrac{1-x\cdot v(x) -2x \cdot v(x)^2}{x^2 + 2x^2 \cdot v(x)}$

I am not sure how to proceed with that, or maybe it is not the right approach for my problem.

Many thanks!

Answer 1

The substitution you did was rather complicated. If we say $v=x+2y$ then we have $v’=1+2y’$ so the differential equation is rewritten as $$\frac{v’-1}{2}=\frac1v$$ which is separable

Answer 2

$$y' = \dfrac{1}{x+2y}$$ Consider $x'$ it's easier as a function of the variable y $$x+2y =x' $$ $$x'-x=2y$$ $$(xe^{-y})'=2ye^{-y}$$ Integrate.. $$(xe^{-y})=2\int ye^{-y} dy$$ $$x(y)=2e^{y}\int ye^{-y} dy$$

Answer 3

You have $$y' = \dfrac{1}{x+2y}$$

That is $$ \frac {dy}{dx} = \dfrac{1}{x+2y}$$

Thus $$ \frac {dx}{dy} = x+2y$$

This is a linear equation and you can solve it for x as a function of y.

Answer 4

hint...You can write this as$$\frac{dx}{dy}=x+2y$$ Or, $$\frac{dx}{dy}-x=2y$$ and this can be solved with an integrating factor $$e^{\int-1dy}$$