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Since f(x, y) = 1 + y2 and "∂f/∂y" = 2y are continuous everywhere, the region r in theorem 1.2.1 can be taken to be the entire xy-plane. use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y(0) = 0.

Respuesta :

Answer:

The solution to the differential equation

y' = 1 + y²

is

y = tan x

Step-by-step explanation:

Given the differential equation

y' = 1 + y²

This can be written as

dy/dx = 1 + y²

Separate the variables

dy/(1 + y²) = dx

Integrate both sides

tan^(-1)y = x + c

y = tan(x+c)

Using the initial condition

y(0) = 0

0 = tan(0 + c)

tan c = 0

c = tan^(-1) 0 = 0

y = tan x

lhmarianateixeira

In this exercise we have to use our knowledge of differential equations to calculate the value of the first solution, so we have to:

[tex]y = tan x[/tex]

Then say the differential equation as:

[tex]y' = 1 + y^2[/tex]

then rewriting as:

[tex]dy/dx = 1 + y^2\\dy/(1 + y^2) = dx[/tex]

Integrate both sides, we have that:

[tex]tan^{(-1)}y = x + c\\y = tan(x+c)[/tex]

So we already have a preview of the solution, so we will have to apply the initial conditions and this results in:

[tex]y(0) = 0\\0 = tan(0 + c)\\tan c = 0\\c = tan^{(-1)} 0 = 0\\y = tan x[/tex]

See more about differential equations at brainly.com/question/2263981