Question

Solve dy/dx = sqrt x+16 subject to the initial condition y(0)=0

Answer 1

Answer:

$y=\frac{2}{3} x^{\frac{3}{2}} + 16x$

or

$y=\frac{2}{3} \sqrt{x^{3}} + 16x$

Step-by-step explanation:

$\frac{dy}{dx} = \sqrt{x} + 16\\ \int\limits {dy} \ =\int\limits { (\sqrt{x} + 16)} \, dx =\\=\int\limits { (x^{\frac{1}{2} } + 16)} \, dx=\int\limits {x^{\frac{1}{2} } } \, dx +\int\limits{16} \, dx = \frac{2}{3} x^{\frac{3}{2}} + 16x + C\\y=\frac{2}{3} x^{\frac{3}{2}} + 16x + C\\0=\frac{2}{3} *0^{\frac{3}{2}} + 16*0 + C\\C=0\\y=\frac{2}{3} x^{\frac{3}{2}} + 16x$