Question

Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y2.

Answer 1

Answer:

The family of solution of the given differential equation is

$y^{l} = 1+y^2$

Step-by-step explanation:

Step 1:-

given equation is y=tan(x+c).......(1)

Differentiating equation (1) with respective to x

$\frac{d y}{d x} = sec^{2} (x+c)$.......(2)

we use trigonometry formula

$sec^{2}x - tan^{2}x=1$

$sec^{2}x=1+tan^{2}x$

step 2 :-

from (2)

$\frac{d y}{d x} = 1+tan^{2}(x+c)$

$y^{1} = 1+y^2$ ( from(1)

the family of solution of the given differential equation.

$y^{1} = 1+y^2$

Answer 2

Answer:

,,,

Step-by-step explanation: