Ytukyg

I've tried the Bernoulli Eq technique, $y=ux$ substitution technique, but nothing seemed to work out. Any ideas?

Put $y=u^2$ so $y'=2uu'=xu^2+uRightarrow u'=frac 12xu+frac12$. Solve for $u$ using integrating factor and then find $y$.

$y'=xy+y^{1/2} implies y'-xy=y^{1/2}$

So taking $p(x)=-x,q(x)=1, n={1/2}$

We have $y'+p(x)=q(x)y^n$

Applying Bernoulli's Equation

$$y^{1-n}=frac{(1-n)int e^{(1-n)int p(x)dx}q(x)dx+c_1}{e^{(1-n)int p(x)dx}}$$

Plugging in some terms we get
$$y^{1/2}=frac{(1/2)int e^{(-x^2/4)}dx+c_1}{e^{(-x^2/4)}}$$

$$y^{1/2}=frac{(sqrt{pi}/2)operatorname{erf}(x/2)+c_1}{e^{(-x^2/4)}}$$

$$y^{1/2}=e^{(x^2/4)}bigg((sqrt{pi}/2)operatorname{erf}(x/2)+c_1 bigg)$$

$$y=frac{pi}{4}e^{(x^2/2)}bigg(operatorname{erf}(x/2)+c_1 bigg)^2$$

Where $operatorname{erf}(z) =frac{2}{sqrt{pi}} int_0^z e^{-t^2} dt $ is the error function

As Guacho Perez answered, starting with $y=u^2$, you end with

$$u left(-2 u'+x u+1right)=0$$ Excluding the trivial case $u=0$, you then need to solve
$$2u'-xu=-1$$ Consider first
$$2u'-xu=0 implies frac{u'}u=frac x 2implies log(u)=frac {x^2} 4+Cimplies u=C e^{frac{x^2}{4}}$$ Now, use the integration factor or the variation of parameters to get
$$2 e^{frac{x^2}{4}} C'(x)=1implies C'=frac 12 e^{-frac{x^2}{4}}implies C=frac{sqrt{pi }}{2} text{erf}left(frac{x}{2}right)+K$$ then $u$ and $y$