Question

sin x. cos y dy + sin y cos x dx = 0.​

Answer 1

EXPLANATION.

⇒ sin x. cos y dy + sin y cos x dx = 0.

As we know that,

We can write equation as,

⇒ sin x .cos y dy = - sin y cos x dx.

⇒ cos y dy/sin y = - cos x dx/sin x.

Integrate both sides of the equation, we get.

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

⇒ ∫cot y dy = - ∫cot x dx.

⇒ ㏑ (sin y ) = - ㏑(sin x) + c.

⇒ ㏑(sin y) + ㏑(sin x) = c.

⇒ sin(y).sin(x) = e^(c).

⇒ sin(y).sin(x) = k.

                                                                                                                         

MORE INFORMATION.

Differential equations of the form of :

dy/dx = f(ax + by + c).

To solve this type of differential equations, we put ax + by + c = v and

dy/dx = 1/b(dv/dx - a)

∴ dv/a + b f(v) = dx.

So, solution is by integrating  :

∫ dv/a + b f(v) = ∫dx.

Answer 2

Step-by-step explanation:

Cos(x).cos(y).dy + sin(x).sin(y) dx = 0

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

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

By taking integration both side,

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

⇒ ∫cot(y) dy = -∫tan(x).dx

⇒ ln|siny| = - (-ln|cos(x)| +c)

⇒ ln|sin(y)| = ln|cos(x)| + ln|C|

⇒ ln|sin(y)| = ln|C.cos(x)| ∵ ln(a) +ln(b) = ln(ab)

⇒ sin(y) = C.cos(x)