Question

solve tany* dy/dx=sin(x+y)+sin(x-y)​

Answer 1

EXPLANATION.

⇒ tan(y).dy/dx = sin(x + y) + sin(x - y).

As we know that,

⇒ dy/dx = sin(x + y) + sin(x - y)/tan(y).

⇒ dy/dx = sin(x)cos(y) + sin(y)cos(x) + sin(x)cos(y) - sin(y)cos(x)/tan(y).

⇒ dy/dx = sin(x)cos(y) + sin(x)cos(y)/tan(y).

⇒ dy/dx = 2 sin(x)cos(y)/tan(y).

⇒ tan(y)/cos(y) dy = 2sin(x)dx.

⇒ sin(y)/cos²(y) dy = 2sin(x)dx.

Integrate both sides, we get.

⇒ ∫sin(y)/cos²(y)dy = ∫2sin(x)dx.

From L.H.S, we get.

By using substitution method, we get.

Let we assume that,

⇒ cos(y) = t.

⇒ Differentiate w.r.t x , we get.

⇒ -sin(y)dy = dt.

Put the value in the equation, we get.

⇒ ∫-dt/t² = 1/t. = 1/cos(y).

⇒ 1/cos(y) = 2(-cos x) + c.

⇒ 1/cos(y) = C - 2cos(x).

                                                                                                                         

MORE INFORMATION.

To solve the homogenous differential equation dy/dx = f(x, y)/g(x, y) , substitute y = vx and So, dy/dx = v + x.dv/dx.

Thus, v + x dy/dx = f(v) = dx/x = dv/f(v) - v.

Therefore solution is,

⇒ ∫dx/x = ∫dv/f(v) - v + c.

Answer 2

$\huge\underline\orange{Given}$

solve tany dy/dx=sin(x+y)+sin(x+y)

$\huge\underline\pink{Explanation}$

→$\frac{dy}{dx}$=

sin(x+y)+sin(x-y)

____________

tany

=[te]\frax{2sinxcosy}{tany}[/tex]

→$\frac{tany}{cosy}$dy=2sin x dx

→[tex]\frac{siny}{cos²ydy}=2sin x dx

integrating we get→

siny

____ = 2sin x dx

cos²ydy

so required solution is →1/cos y=c-2 cos