Hey,
Could anyone help me with this calculus question?
Find y in terms of x
dy/dx = √x+5 / √x
$\displaystyle \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}}$
$\displaystyle dy = \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx$
$\displaystyle \int y = \int \left( \sqrt{x} + \frac{5}{\sqrt{x}}\right) dx$
$\displaystyle y = \frac{2}{3}(x)^{\frac{3}{2}} + 2(5 \sqrt{x})$
yeah I forgot to add the constant but
your question is
$\displaystyle \frac{dy}{dx} = \frac{\sqrt{x}+5}{\sqrt{x}}$ not
$\displaystyle \frac{dy}{dx} = \sqrt{x} + \frac{5}{\sqrt{x}} $
I solved the second one the first is the same , you should just integrate $\displaystyle 1 + \frac{5}{\sqrt{x}} $ instead of $\displaystyle \sqrt{x} + \frac{5}{\sqrt{x}}$
that's your fault you should put bracket like this dy/dx = (sqrt(x) + 5)/sqrt(x)
$\displaystyle \frac{dy}{dx} = \frac{\sqrt{x}+5}{\sqrt{x}}$
$\displaystyle \int dy= \int \left( \frac{\sqrt{x}+5}{\sqrt{x}}\right) dx$
$\displaystyle y = \int \left(1 + \frac{5}{\sqrt{x}}\right) dx$
$\displaystyle y = x + \frac{5\sqrt{x}}{\frac{1}{2}} + c$
$\displaystyle y = x + 10\sqrt{x} + c$