Respuesta :
Answer:
Step-by-step explanation:
a) We can state that the value of y' is negative on an interval because y^2 is positive in any interval (except in y(0), in which y'(0)=0). If y' is negative, as it is the slope of the tangent of the function y in any point, we can conclude that y is decreaing or equal to zero on any interval on which is defined.
b) First, we calculate y'
[tex]y=\frac{4}{x+C}\\\\y'=4*\frac{(-1+0)-0}{(x+C)^2}=\frac{-4}{(x+C)^2}[/tex]
Then we replace in the differential equation and get the same result
[tex]y'=-\frac{1}{4} y^2=-\frac{1}{4} (\frac{4}{x+C} )^2=-\frac{1}{4}\frac{16}{(x+C)^2}=\frac{-4}{(x+C)^2}[/tex]
c) The question is not readable, so I am going to solve this initial-value problem:
[tex]y'=-\frac{1}{4}y^2\\\\y(0)=4[/tex]
[tex]dy/dx=-(1/4)y^2\\\\ \int \frac{dy}{y^2}=-(1/4)\int dx\\\\-\frac{1}{y} =-(1/4)(x+C)\\\\y=\frac{4}{x+C}\\\\y(0)=4=\frac{4}{0+C} \\\\C=1\\\\\\y=\frac{4}{x+1}[/tex]
We can observe sign of rate of function to tell whether its growing or not.
Answers:
A): 1. The function y must be decreasing (or equal to 0) on any interval on which it is defined.
B): Verified below.
C): Solution of the problem [tex]3\sqrt{12}x+ 0.008 = 12[/tex] is x = 1.154
A): Analysis of solution by seeing differential equation:
Given differential equation is: [tex]y' = -\dfrac{1}{4} y^2[/tex]
How to deduce the results just by seeing it?
The equation tells us that:
rate = negative of ( [tex]\dfrac{1}{2} \times (y^2)[/tex] )
rate = negative of (positive or zero) = negative or zero
Thus, rate is negative or zero no matter what value we put in the place of y from its valid domain, since [tex]y^2 \geq 0[/tex].
When rate is negative or zero, that means the function will never grow upwards. Thus, either decreasing or staying at same level.
Thus, "Option 1: The function y must be decreasing (or equal to 0) on any interval on which it is defined." is correct.
B): Solving the differential equation:
The solution of [tex]y' = -\dfrac{1}{4} y^2[/tex] is derived as:
[tex]\dfrac{dy}{dx} = -\dfrac{1}{4}y^2\\\\-\dfrac{4dy}{y^2} = dx\\[/tex]
Integrating both sides, we get:
[tex]\int -\dfrac{4dy}{y^2} = \int dx\\\\\dfrac{4}{y} = x + C\\y = \dfrac{4}{x + C}[/tex]
Thus, it is verified that all members of the family y = 4/(x + C) are solutions of the equation in part (a).
C):
The solution to the problem [tex]3\sqrt{12}x+ 0.008 = 12[/tex] is found by:
[tex]3\sqrt{12}x+ 0.008 = 12\\\\x = \dfrac{12 - 0.008}{3\sqrt{12}}\\\\x = \dfrac{11.992}{10.392} = 1.154[/tex]
Learn more about differential equations here:
https://brainly.com/question/25731911
