Solving a Differential Equation by Solutions by Substitution

Question

The given equation is $$\frac{dy}{dx}=\sqrt{x+y}$$ I saw it as a separation of variables but decided to solve it as a substitution. At first I though maybe I should just invert the whole thing and try solving it that way, but that got me up to: $$x\frac{dx}{du}=\frac{1}{\sqrt{x+ux}}-\frac{u(\sqrt{u+ux})}{\sqrt{x+ux}}$$ So then I decied to try it as is: $$u+x\frac{du}{dx}=\sqrt{x+ux}$$ $$x\frac{du}{dx}=\sqrt{x+ux}-u$$ I multiplied by the square root because thats usually what I need to do when I subtract $u$ $$x\frac{du}{dx}=\sqrt{x+ux}-\frac{u(\sqrt{x+ux})}{\sqrt{x+ux}}$$ $$x\frac{du}{dx}=\sqrt{x+ux}-\frac{u(x(\sqrt{1+u}))}{x\sqrt{1+u}}$$ $$x\frac{du}{dx}=\sqrt{x+ux}-\frac{u(\sqrt{1+u})}{\sqrt{1+u}}$$ But I'm honestly stuck on how to proceed, should I multiply out the $U$ into the square root(if so how?), which one equation has the path of least frustration and, how can I further simplify the equation with the test I have taken?

Answer 1

Using the new dependent variable $u = x + y$, the differential equation becomes $$ \dfrac{du}{dx} = \sqrt{u}+1$$ which is separable.

Answer 2

The substitution $y=ux$ makes things complicated because the differential equation is not homogeneous. You can choose a much better substitution that makes things easy. $$\frac{dy}{dx}=\sqrt{x+y}$$ Sustitute $$u=\sqrt{x+y} \implies u^2 =x+y$$ implicit differentiation gives: $$2uu'=1+y' \implies y'=2uu'-1$$ The equation becomes: $$2uu'-1=u$$ It's separable. $$\dfrac {u }{1+u}\; du=\frac 12dx$$