Answer:
The solution of differential equation =
y = [tex]\boldsymbol{\frac{10}{.2t - 1}}[/tex]
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
[tex]\boldsymbol{\frac{\mathrm{d} y}{\mathrm{d} t} = -0.02 y^{2}}[/tex]
[tex]\frac{\mathrm{d} y}{y^{2}} = -0.02\mathrm{d} t[/tex]
Integrate both side
[tex]\int \frac{\mathrm{d} y}{y^{2}} = \int -0.02\mathrm{d} t[/tex]
[tex]\frac{y^{-2 + 1}}{-2 + 1} = -.02\times t[/tex] + C
[tex]\frac{- 1 }{y} = -.02\times t + C[/tex]
when t = 0 then y = 10
[tex]\frac{- 1 }{10} = -.02\times 0 + C[/tex]
C = [tex]\frac{- 1 }{10}[/tex]
[tex]\frac{- 1 }{y} = -.02\times t - \frac{ 1 }{10}[/tex]
[tex]\frac{ 1 }{y} = .02\times t - \frac{ 1 }{10}[/tex]
[tex]\frac{ 1 }{y} = \frac{.2t - 1}{10}[/tex]
y = [tex]\frac{10}{.2t - 1}[/tex]
The solution of the differential equation is:
[tex]y = \frac{1}{-0.1 + 0.02*t}[/tex]
Here we need to solve:
[tex]\frac{dy}{dt} = -0.02*y^2[/tex]
This is a separable differential equation, we can rewrite this as:
[tex]\frac{dy}{y^2} = -0.02dt[/tex]
Now we integrate in both sides to get:
[tex]\int\limits\frac{dy}{y^2} = \int\limits-0.02dt\\\\-\frac{1}{y} = -0.02*t + C[/tex]
Where C is a constant of integration.
Solving for y, we get:
[tex]y = \frac{1}{-C + 0.02*t}[/tex]
And we know that for t = 0, there are 10 grams of substance, then:
[tex]10 = \frac{1}{-C + 0.02*0} = -\frac{1}{C} \\\\C = -1/10 = -0.1[/tex]
So the equation is:
[tex]y = \frac{1}{-0.1 + 0.02*t}[/tex]
If you want to learn more about differential equations, you can read:
https://brainly.com/question/18760518
