Question

(a) What can you say about a solution of the equation y' = - (1/4)y2 just by looking at the differential equation? 1. The function y must be decreasing (or equal to 0) on any interval on which it is defined. 2. The function y must be increasing (or equal to 0) on any interval on which it is defined. (b) Verify that all members of the family y = 4/(x + C) are solutions of the equation in part (a). (c) Find a solution of the initial-value problem. 3√12x+ 0.008 12​

Answer 1

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'

$y=\frac{4}{x+C}\\\\y'=4*\frac{(-1+0)-0}{(x+C)^2}=\frac{-4}{(x+C)^2}$

Then we replace in the differential equation and get the same result

$y'=-\frac{1}{4} y^2=-\frac{1}{4} (\frac{4}{x+C} )^2=-\frac{1}{4}\frac{16}{(x+C)^2}=\frac{-4}{(x+C)^2}$

c) The question is not readable, so I am going to solve this initial-value problem:

$y'=-\frac{1}{4}y^2\\\\y(0)=4$

$dy/dx=-(1/4)y^2\\\\ \int \frac{dy}{y^2}=-(1/4)\int dx\\\\-\frac{1}{y} =-(1/4)(x+C)\\\\y=\frac{4}{x+C}\\\\y(0)=4=\frac{4}{0+C} \\\\C=1\\\\\\y=\frac{4}{x+1}$