How to solve the DE $y' = -y + ty^{1/2}$? The DE is $y' = -y + ty^{\frac{1}{2}}$. 
$2 \le t \le 3$
$y(2) = 2$
I tried to see if it was in the linear form. I got:
$$\frac{dy}{dt} + y = ty^{\frac{1}{2}}$$
The RHS was not a function of t. I also tried separation of variables, but I couldn't isolate the y from the term $ty^{\frac{1}{2}}$. Any hints?
 A: Set $y=z^2$ and simplify. You get $y'=2zz'$ and your equation is $2zz'+z^2=tz$ or $2z'+z=t$ which is linear and you can apply integrating factor to it.
(We assumed $y>0$, which it is near initial point.)
A: Let $y=u^2$, then you can cancel a factor of $u$ and get
$$2 u' + u = t$$
for which you can apply an integrating factor of $e^{t/2}$ to both sides and get
$$\frac{d}{dt} [u e^{t/2}] = \frac{t}{2} e^{t/2}$$
Integrating both sides, we get the general solution:
$$u(t) = t-2 + C e^{-t/2}$$
where $C$ is a constant of integration.  The solution is then $y(t)=u(t)^2$.
A: I'll just add that integration factor is not necessary. From the equations
$$
2z'+z = t
$$
you can assume that particular solution is linear $z^p = At+B$ and substitute it to ODE
$$
2A+At+B=t
$$
from which you can easily find that $A = 1, B = -2$, so $z^p = t - 2$. General solution of inhomogeneous problem is a sum of general solution of homogeneous problem and particular solution of inhomogeneous problem. Homogeneous one can be easily solved and $z_0 = C^{-\frac 12t}$, so $z = t-2+C^{-\frac 12 t}$
A: If that $y^{\frac12}$ weren't there, you might solve the problem by multiplying by an integrating factor of $e^t$ to yield $(e^ty)'$ on the left side.  Try making the substitution
$$z=e^ty$$
$$y^{\frac12}=z^{\frac12}e^{-\frac t2}$$
$$e^ty'+e^ty=(e^ty)'=te^ty^{\frac12}$$
$$z'=te^t(z^{\frac12}e^{-\frac t2})=te^{\frac t2}z^{\frac12}$$
This equation is now separable.
A: Hint: 
Another approach applying here is to see that your OD is a [Bernoulli] (http://en.wikipedia.org/wiki/Bernoulli_differential_equation) equation. Just to see that $n=1/2$. The substitution $w=y^{1-1/2}=\sqrt{y}$ works.
