Question

4.Solution of the differential equation
sec)^2 xtanydx+ ( sec]^2 ytanxdy=0 is
Is​

Answer 1

EXPLANATION.

⇒ sec²x tan y dx + sec²y tan x dy = 0.

As we know that,

We can write equation as,

⇒ sec²y tan x dy = - sec²x tan y dx.

⇒ sec²y dy/tan y = - sec²x dx/tan x.

Integrate both sides of the equation, we get.

⇒ ∫(sec²ydy)/(tan y) = - ∫(sec²xdx)/(tan x).

By using substitution method, we get.

Let we assume that,

⇒ tan y = t.

Differentiate w.r.t y, we get.

⇒ sec²ydy = dt.

⇒ tan x = z.

Differentiate w.r.t x, we get.

⇒ sec²xdx = dz.

Put the values in the equation, we get.

⇒ ∫dt/t = - ∫dz/z.

⇒ ㏑|t| = - ㏑|z| + C.

Put the value of t = tan y and z = tan x in the equation, we get.

⇒ ㏑|tan y| = - ㏑|tan x| + C.

⇒ ㏑|tan y| + ㏑|tan x| = C.

                                                                                                                   

MORE INFORMATION.

Differential equation of Homogenous type Or f(y/x) = 0.

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 dv/dx = f(v) ⇒ dx/x = dv/f(v) - v.

Therefore solution is ∫dx/x = ∫dv/f(v) - v + C.