Math, asked by oluwaariyooa, 7 months ago

solve the equation sin y-cos (y+38)=0​

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Answered by JasynAlexanderIII
1

Answer:

Here u go hope this helps

Step-by-step explanation:

2.2 Separable Equations

An equation y

0 = f(x, y) is called separable provided algebraic operations, usually multiplication, division and factorization, allow it to be

written in a separable form y

0 = F(x)G(y) for some functions F and

G. This class includes the quadrature equations y

0 = F(x). Separable

equations and associated solution methods were discovered by G. Leibniz

in 1691 and formalized by J. Bernoulli in 1694.

Finding a Separable Form

Given differential equation y

0 = f(x, y), invent values x0, y0 such that

f(x0, y0) 6= 0. Define F, G by the formulas

F(x) = f(x, y0)

f(x0, y0)

(1) , G(y) = f(x0, y).

Because f(x0, y0) 6= 0, then (1) makes sense. Test I below implies the

following test.

Theorem 2 (Separability Test)

Let F and G be defined by equation (1). Compute F(x)G(y). Then

(a) F(x)G(y) = f(x, y) implies y

0 = f(x, y) is separable.

(b) F(x)G(y) 6= f(x, y) implies y

0 = f(x, y) is not separable.

Invention and Application. Initially, let (x0, y0) be (0, 0) or (1, 1)

or some suitable pair, for which f(x0, y0) 6= 0; then define F and G by

(1). Multiply to test the equation F G = f. The algebra will discover a

factorization f = F(x)G(y) without having to know algebraic tricks like

factorizing multi-variable equations. But if F G 6= f, then the algebra

proves the equation is not separable.

Non-Separability Tests

Test I Equation y

0 = f(x, y) is not separable if for some pair of

points (x0, y0), (x, y) in the domain of f

(2) f(x, y0)f(x0, y) − f(x0, y0)f(x, y) 6= 0.

Test II The equation y

0 = f(x, y) is not separable if either

fx(x, y)/f(x, y) is non-constant in y or fy(x, y)/f(x, y)

is non-constant in x.

Illustration. Consider y

0 = xy + y

2

. Test I implies it is not separable,

because f(x, 1)f(0, y)−f(0, 1)f(x, y) = (x+1)y

2−(xy+y

2

) = x(y

2−y) 6=

0. Test II implies it is not separable, because fx/f = 1/(x + y) is not

constant as a function of y.

74

Test I details. Assume f(x, y) = F(x)G(y), then equation (2) fails because each term on the left side of (2) evaluates to F(x)G(y0)F(x0)G(y)

for all choices of (x0, y0) and (x, y) (hence contradiction 0 6= 0).

Test II details. Assume f(x, y) = F(x)G(y) and F, G are sufficiently

differentiable. Then fx(x, y)/f(x, y) = F

0

(x)/F(x) is independent of y

and fy(x, y)/f(x, y) = G0

(y)/G(y) is independent of x.

Separated Form Test

A separated equation y

0/G(y) = F(x) is recognized by this test:

Left Side Test. The left side of the equation has factor y

0

and it is independent of symbol x.

Right Side Test. The right side of the equation is independent of symbols y and y

0

.

Variables-Separable Method

Determined by the method are the following kinds of solution formulas.

Equilibrium Solutions. They are the constant solutions y = c of y

0 =

f(x, y). Find them by substituting y = c in y

0 = f(x, y), followed

by solving for c, then report the list of answers y = c for all values

of c.

Non-Equilibrium Solutions. For separable equation y

0 = F(x)G(y),

it is a solution y with G(y) 6= 0. It is found by dividing by G(y)

and applying the method of quadrature.

The term equilibrium is borrowed from kinematics. Alternative terms

are rest solution and stationary solution; all mean y

0 = 0 in calculus

terms.

Spurious Solutions. If F(x)G(y) = 0 is solved instead of G(y) =

0, then both x and y solutions might be found. The x-solutions are

ignored: they are not equilibrium solutions. Only solutions of the form

y = constant are called equilibrium solutions.

It is important to check the solution to a separable equation, because

certain steps used to arrive at the solution may not be reversible.

For most applications, the two kinds of solutions suffice to determine all

possible solutions. In general, a separable equation may have non-unique

solutions to some initial value problem. To prevent this from happening,

it can be assumed that F, G and G0 are continuous; see the PicardLindel¨of theorem, page 61. If non-uniqueness does occur, then often

the equilibrium and non-equilibrium solutions can be pieced together to

represent all solutions.

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