Finding solution to linear equations using integrating factors Here's a couple of problems:


*

*$$2xy' + y = 2 \sqrt{x}$$
so according to my book, I should get it into this form:

so let's divide by 2x:
$$y' + \frac{1}{2x} * y = \frac{2x^{\frac{1}{2}}}{2x} = x^{\frac{-1}{2}}$$
So is the integrating factor = $$I(x) = e^{\int \frac{1}{2x}} = e^{\frac{1}{2} ln(2x)}$$
A bit stuck from here. How does that integral reduce?
EDIT
going further:
$$I(x) = \sqrt{2x}$$
so multiplying both sides:
$$\sqrt{2x} \frac{dy}{dx} + \frac{1}{\sqrt{2x}} y = \sqrt{2x}x^{\frac{-1}{2}} = \sqrt{2}$$
$$(\sqrt{2x}y)' = \sqrt{2}$$
$$\sqrt{2x}y = \sqrt{2}x + c$$
$$y = \frac{\sqrt{2}x + c}{\sqrt{2x}} = x^{\frac{1}{2}} + \frac{c}{\sqrt{2x}}$$
When plugging this back in... It's not equal. Did I do something wrong?


*$$xy' + y = \sqrt{x}$$
divide by x:
$$\frac{dy}{dx} + \frac{1}{x}y = x^{\frac{-1}{2}}$$
so the integrating factor is: $e^{\int \frac{1}{x}dx} = x$
$$x * \frac{dy}{dx} + y = x^{\frac{1}{2}}$$
$$(xy)' = x^{\frac{1}{2}}$$
$$xy = \frac{2}{3} x^{\frac{3}{2}}$$
$$y = \frac{2}{3} x^{\frac{1}{2}}$$
Is that right?
 A: *

*$$I(x)=e^{\frac12\ln(2x)}=e^{\ln(\sqrt{2x})}=\sqrt{2x}$$ Note that $\sqrt2$ is just a constant and won't affect your solutions if you multiply the equation through by it, so you can ignore it, and instead take $$I(x)=\sqrt x$$ Another way to see why this is ok is that in computing $\int\frac1{2x}dx$, you can simply do this: $$=\int\frac12 \frac1x dx=\frac12\ln x$$

*You have the right idea, but when you integrated $(xy)'=x^{1/2}$, you forgot about the constant of integration. So you should get $$xy=\frac23x^{3/2}+c\\y=\frac23x^{1/2}+\frac cx$$

Edit: 
You said "plugging this in, it didn't work". Are you sure? Your solution is correct. By the way, if you let $c=\sqrt2C$, $$\frac{c}{\sqrt{2x}}=\frac{C}{\sqrt x}$$which may make differentiation easier (so you don't have to keep using chain rule). With the way you currently have the solution, here is how you'd go about plugging it back in: 
$$y=x^{1/2}+x(2x)^{-1/2}\\y'=\frac12x^{-1/2}+c(2)(-1/2)(2x)^{-3/2}=\frac12x^{-1/2}-\frac c{(2x)^{3/2}}\\\implies 2xy'=\sqrt x-\frac c{\sqrt{2x}}\\\implies 2xy'+y=2\sqrt x$$Which is correct.
