Solution of the differential equation tany sec2x dx + tan x sec2ydy = 0 is

Question

Solution of the differential equation tany sec2x dx + tan x sec2ydy = 0 is (A) tan x + tan y = k (B) tan x- tan y - k (C) (tan x/tan y)=k (D) tan x⋅ tan y = k

Tardigrade Admin · Accepted Answer

tan y sec2xdx+tanx sec2ydy=0 ⇒ (sec2 xdx/tan x)=-(sec2 ydy/tan y) On integration, we get log tanx = - log tany + log k ⇒ log (tan x⋅ tan y) = log k ⇒ tanx⋅ tany=k

Q. Solution of the differential equation
$t an y se c^{2} x d x + t an x se c^{2} y d y = 0$ is

$t an x + t an y = k$

$t an x - t an y - k$

$\frac{t an x}{t an y} = k$

$t an x \cdot t an y = k$

$t an y se c^{2} x d x + t an x se c^{2} y d y = 0$
$\Rightarrow \frac{se c ^{2} x d x}{t an x} = - \frac{se c ^{2} y d y}{t an y}$
On integration, we get $l o g t an x = - l o g t an y + l o g k$
$\Rightarrow l o g (t an x \cdot t an y) = l o g k$
$\Rightarrow t an x \cdot t an y = k$

Q. Solution of the differential equation tanysec2xdx+tanxsec2ydy=0 is

tanx+tany=k

tanx−tany−k

tanytanx​=k

tanx⋅tany=k