Can anyone help with ODE?!!!!!

Can one one help me with these three problems:
1)Solve the equation y'=x^2y^(1/3) (Don't forget the singular solution)

2)Solve the IVP y'=e^(x+y) , y(0)=0

3)Verify that f(x)=(-x/2)-(1/4)+ce^(2x) is a solution of the equation y'=2y+x on (-inf,inf). Find a solution whose graph passes through the point (-1,1/2).

Any help would be great, thanks a bunch

The first two are variables separable differential equations.

1. The singular solution is y = 0 because this makes both sides of the differential equation 0.

dy/dx = x^2*y^(1/3) ----> INT y^(-1/3) dy = INT x^2 dx

(3/2)*y^(2/3) = (1/3)*x^3 + c ----> (1/27)*(2x^3 + k)^(3/2) ....... k = 2c

2. dy/dx = e^(x + y) = (e^x)(e^y) ----> INT e^-y dy = INT e^x dx

----> - (e^-y) = e^x + c

y(0) = 0 ----> -1 = 1 + c ----> c = -2 ----> - (e^-y) = e^x - 2

e^-y = 2 - e^x ----> -y = ln(2 - e^x)

----> y = - ln(2 - e^x) .............. (Note that domain is x < ln2)

3. Differentiate the given function to get f '(x) and then replace y by f(x) and y' by f '(x) and show that the differential equation works.

x = -1, y = 1/2 substituted in the given function shows c = (e^2)/4

1) dy/dx = x^2y^(1/3) so dy/(y^(1/3) = (x^2)dx, or [y^(2/3)]/(2/3) = (x^3)/3 + c.
There is no singular solution for this.

2) Let x+y = t so dy/dx = dt/dx -1 and si dt/dx - 1 = e^t
so dt/dx = 1+e^t
dt/(1+e^t) = dx
e^(-t)/[e^(-t)+1] = dx.
Integrating we get,
-ln([e^(-t)+1] = x+c.
-ln[e^(-x-y)+1]=x+c.
Now y(0) = 0 means y = 0 when x = 0, in case you donot know.

so -ln2 = x
hence x-ln2 = -ln[(e^(-x-y)+1] is the P. S.
3) First part is practicing differentitation. In second part put x= -1 and y = 1/2 and find c. Rewrite the eqn with this value of c.