Question

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

(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y².
1) Differentiating y = tan(x+c) we get y'=1+sec, the solution is not defined on (−2, 2).
2) Differentiating y = tan(x+c) we get y'=tan²(x+c) or y'=1+y².
3) Differentiating y = tan(x+c) we get y'=sec(x+c) or y'=1+y².
4) Differentiating y = tan(x+c) we get y'=1+tan²(x+c) or y'=1+y².
5) Differentiating y = tan(x+c) we get y'=csc(x+c) or y'=1+y².
(b) 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.
y=_____.
Even though x0 = 0 is in the interval (−2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = ±(_____) , the solution is not defined on (−2, 2).
(c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). (Enter your answer using interval notation.)

Answer 1

The solution y = tan(x + c) is a one-parameter family of solutions to the differential equation y' = 1 + y². The specific initial-value problem y' = 1 + y², y(0) = 0 has the explicit solution y = tan(x). The solution y = tan(x) has its largest interval of definition as (-π/2, π/2) to avoid the discontinuities of the tangent function.

Step-by-step explanation:

To verify that y = tan(x + c) is a solution to the differential equation y' = 1 + y², we must differentiate y with respect to x and check if the resultant expression matches the given differential equation. Using the chain rule for differentiation, we have:

y' = sec²(x+c)

Since tan²(x+c) = sec²(x+c) - 1, we can rewrite y' as:

y' = 1 + tan²(x+c)
y' = 1 + y²

Therefore, y = tan(x + c) is indeed a one-parameter family of solutions to the differential equation y' = 1 + y².

To solve the initial-value problem y' = 1 + y², y(0) = 0, we look for a specific value of c that satisfies the initial condition. Since y is the tangent function, y(0) = 0 implies that the argument inside tangent must be an integer multiple of π. Thus, c must be 0, making the explicit solution y = tan(x).

Tan(x) has discontinuities at x = ±(π/2 + nπ), where n is an integer. Therefore, the intervals of definition for y avoid these values.

The largest interval of definition for the solution that includes x = 0, without crossing discontinuities, is (-π/2, π/2).