Question

This is a Differential Equations problem, I only need help on part 1 from question five. I need steps as well, thank you.

Answer 1

Answer:

a) $y(t) = \sqrt{2t + 1 }$

b) 1.5 hours after thickness will be 2 inches.

Step-by-step explanation:

$\frac{dy}{dt} = \frac{1}{y} \\ \\ ydy = dt \\ \\ integrating \: both \: sides \\ \\ \int y \: dy = \int 1 \: dt \\ \\ \frac{ {y}^{2} }{2} = t + c \\ \\ {y}^{2} = 2t + 2c \\ \\ y (t)= \sqrt{2t + 2c} ...(1) \\ \\ a) \: \: \: plug \: t = 0 \: in \: (1) \\ \\ y (0)= \sqrt{2(0) +2 c} \\ \\ y (0)= \sqrt{0 +2 c} \\ \\ 1 = \sqrt{2c} \: \: ( \because \: y (0)=1) \\ \\ 2c = 1 \: \implies \: c = \frac{1}{2} \\ \\ plug \: c = \frac{1}{2} \: in \: (1) \\ \\ y(t) = \sqrt{2t + 2 \times \frac{1}{2} } \\ \\ \huge \: \red {y(t) = \sqrt{2t + 1 } } \\ \\b) \: \: plug \: y(t) = 2 \: in \: above \: equation \\ \\ 2 = \sqrt{2t + 1} \\ \\ 4 = 2t + 1 \: \\ (squaring \: both \: sides) \\ \\ 4 - 1 = 2t \\ \\ 2t = 3 \\ \\ t = \frac{3}{2} \\ \\ t = 1.5 \: hours$