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

Answer 2

The solution of the differential equation is:

$y = \frac{1}{-0.1 + 0.02*t}$

How to solve the differential equation?

Here we need to solve:

$\frac{dy}{dt} = -0.02*y^2$

This is a separable differential equation, we can rewrite this as:

$\frac{dy}{y^2} = -0.02dt$

Now we integrate in both sides to get:

$\int\limits\frac{dy}{y^2} = \int\limits-0.02dt\\\\-\frac{1}{y} = -0.02*t + C$

Where C is a constant of integration.

Solving for y, we get:

$y = \frac{1}{-C + 0.02*t}$

And we know that for t = 0, there are 10 grams of substance, then:

$10 = \frac{1}{-C + 0.02*0} = -\frac{1}{C} \\\\C = -1/10 = -0.1$

So the equation is:

$y = \frac{1}{-0.1 + 0.02*t}$

If you want to learn more about differential equations, you can read:

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