Question

During a chemical reaction, the function y=f(t) models the amount of a substance present, in grams, at time t seconds. At the start of the reaction (t=0) , there are 10 grams of the substance present. The function y=f(t) satisfies the differential equation dydt=−0.02y^2

Answer 1

Answer:

The solution of differential equation =

y = $\boldsymbol{\frac{10}{.2t - 1}}$

Step-by-step explanation:

Given-

y=f(t) models the amount of a substance present, in grams, at time t seconds

when t = 0  then y = 10

the differential equation is

$\boldsymbol{\frac{\mathrm{d} y}{\mathrm{d} t} = -0.02 y^{2}}$

$\frac{\mathrm{d} y}{y^{2}} = -0.02\mathrm{d} t$

Integrate both side

$\int \frac{\mathrm{d} y}{y^{2}} = \int -0.02\mathrm{d} t$

$\frac{y^{-2 + 1}}{-2 + 1} = -.02\times t$ + C

$\frac{- 1 }{y} = -.02\times t + C$

when t = 0  then y = 10

$\frac{- 1 }{10} = -.02\times 0 + C$

C = $\frac{- 1 }{10}$

$\frac{- 1 }{y} = -.02\times t - \frac{ 1 }{10}$

$\frac{ 1 }{y} = .02\times t - \frac{ 1 }{10}$

$\frac{ 1 }{y} = \frac{.2t - 1}{10}$

y = $\frac{10}{.2t - 1}$