Question

Find the solution of $(3\tan x+4\cot y-7){\sin }^{2}ydx$ $-(4\tan x+7\cot y-5){\cos }^{2}xdy=0$.

• A. $3\frac{{\tan }^{2}x}{2}-7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$
• B. $8\frac{{\tan }^{2}x}{2}-7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$
• C. $3\frac{{\tan }^{2}x}{2}+7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$
• D. $8\frac{{\tan }^{2}x}{2}+7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$

Answer 1

The correct option is A $3\frac{{\tan }^{2}x}{2}-7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$

Given $(3\tan x+4\cot y-7){\sin }^{2}ydx-(4\tan x+7\cot y-5){\cos }^{2}xdy=0$
Dividing throughout by ${\sin }^{2}y{\cos }^{2}x,$ we get

$3\tan x{\sec }^{2}xdx-7{\sec }^{2}dx-7\cot ycose{c}^{2}ydy+5cose{c}^{2}ydy+4(\cot y{\sec }^{2}xdx-\tan xcose{c}^{2}y)dy=0$

Integrating on both sides

$3\int \tan x{\sec }^{2}xdx-7\int {\sec }^{2}dx-7\int \cot ycose{c}^{2}ydy+5\int cose{c}^{2}ydy$

$+4\int (\cot y{\sec }^{2}xdx-\tan xcose{c}^{2}y)dy=0$

Put $\tan x=u\Rightarrow {\sec }^{2}xdx=du$ for first integral

and $\cot y=v\Rightarrow -cose{c}^{2}ydy=dv$ for third integral
$\therefore 3\int udu-7\tan x-7\int vdv-5\cot y+4\int (\cot y{\sec }^{2}xdx-\tan xcose{c}^{2}y)dy=0$
Since, differentiation of $\cot y\tan x=\cot y{\sec }^{2}dx-\tan xcose{c}^{2}y$
$3\frac{{\tan }^{2}x}{2}-7\tan x+7\frac{{\cot }^{2}y}{2}-5\cot y+4\left(\cot y\tan x\right)=c$