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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 :

abidemiokin

The explicit solution to the differential equation is y = tan x

Differential equation

Given the differential function:

[tex]\frac{dy}{dx} =1 + y^2\\[/tex]

Using the variable separable method;

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

Integrate both sides of the equation

[tex]x = tan^{-1}y \\tan^{-1}y= x+C[/tex]

[tex]y=tan(x +C)[/tex]

Using the initial condition y(0) = 0

If x = 0

0 = tan(0 + C)

0 = tanC

C = arctan 0

C = 0

Recall that y = tan (x+C)

y = tan(x+0)
y = tan x

Hence the explicit solution to the differential equation is y = tan x

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