Question

Show that an implicit solution of 2x sin? (7) dx - (x2 + 14) cos(y) dy = 0 is given by In(x²+ 14) + cse(Y) = C. 2x Differentiating In(x² + 14) + csc ) = we get x²+ 14 + dy = 0 or 2x siny) dx + dx dy = 0. Find the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an arbitrary integer.) y =?

Answer 1

The implicit solution of the given differential equation is In(x²+ 14) + csc(y) = C. There are no constant solutions that were lost in the solution of the differential equation.

Step-by-step explanation:

To show that the implicit solution of the differential equation is given by In(x²+ 14) + csc(y) = C, we need to differentiate the equation and verify that it satisfies the differential equation.

First, we differentiate In(x²+ 14) + csc(y) = C with respect to x:

d/dx(In(x²+ 14)) + d/dx(csc(y)) = 0

2x/(x²+ 14) - csc(y) cot(y) dy/dx = 0

Simplifying, we get:

2x sin(y) dx - (x²+ 14) cos(y) dy = 0

This is the same as the given differential equation, so the implicit solution is verified.

To find the constant solutions that were lost in the solution of the differential equation, we can set (x²+ 14) + cot(y) = 0. Rearranging, we get:

x²+ 14 = -cot(y)

x²+ 14 = -1/tan(y)

x²+ 14 = -tan(y)

tan(y) = -x² - 14

Since tan(y) is periodic with a period of π, we can find the constant solutions by setting -x² - 14 equal to kπ, where k is an arbitrary integer:

x² = -14 - kπ

There are no real solutions for x, so there are no constant solutions.