Unsolvable Linear First-Order ODEs with Boundary Conditions

[lol_nl]

Homework Statement

Out of a set of differential equations with boundary conditions, there are three (first order) equations I couldn't solve. These are:

Homework Equations

1. $$\frac {dy} {dx} = \sqrt{x + y}, y(1) = 0.$$
2. $$\frac {dy} {dx} = 2y(x \sqrt{y} - 1), y(0) = 1.$$
3. $$2x^2 \frac {dy} {dx} = x^2 + y^2, y(2) = 4.$$

The Attempt at a Solution

The first two can probably be solved with a nice substitution. I tried u = x + y for the first one, but this gave me the equation $$\frac {du} {dx} = 1 + \sqrt{u}$$, which can be solved for u to get, after resubstituting, $$2 \sqrt{x+y} - 2 log[1+\sqrt{x+y} - 2 + 2 log[2] = x$$, which doesn't seem solvable for y[x].
For the second one I tried substituting $$u = x \sqrt{y}$$ and $$u = x \sqrt{y} - 1$$, but neither gave an equation that could be written in terms of u only (without x or y).
For the third one I only noticed that y=x is a general solution, but it doesn't agree with the initial condition y(2) = 4, and neither does any manipulation such as y=2x or y=x+2.

 

[lol_nl]

By the way, Mathematica could only solve for (2.) and gave the solution:
$$y[x] = \frac {1}{(1+x+e^{x} C[1])^{2}}$$.

For the first one it couldn't find a solution and for the third one it gave an output of a couple of dozen of lines of amongst other Bessel functions.

Another method I tried was writing the equation in the form
$$a(x,y) + b(x,y) \frac{dy}{dx} = 0$$ and then trying to find a constant E(x,y) such that $$\frac{\partial E}{\partial x} = a(x,y), \frac{\partial E}{\partial y} = b(x,y)$$. However, this requires equality of mixed second-order derivatives, and this is not the case for either of the equations. Neither could I easily see a constant $$\phi(x,y)$$ to multiply the above equation with such that $$\phi a(x,y)$$ and $$\phi b(x,y)$$ do satisfy.
(I'm not sure what this method is called in English; it was mentioned in my Dutch textbook)

 

[Bohrok]

The first one can't be solved for y with elementary functions, so that's about as far as you can go.

Apparently the second one can be written as a Bernoulli equation according to WolframAlpha, and that's how you would get the solution that Mathematica gave you.

For the third one, y = ux will turn it into a separable equation.
I think it's interesting how x works in the differential equation, but you can't get it from the general solution...

The method you're talking about is exact differential equations I think; it doesn't look like that method will be helpful for this one.