sin x. cos y dy + sin y cos x dx = 0.
Answers
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.
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)