Question

Solution of differential equation $\frac{dx}{dy}=\tan x(1+ysinx)$ is given by -

• A. $cosecx=-y+1+{Ce}^{-y}$
• B. $y=\tan x+{Ce}^X$
• C. $\sin x{e}^y=1+y+C$
• D. $cosecx=y+{Ce}^y$

Answer 1

The correct option is A $cosecx=-y+1+{Ce}^{-y}$

$\cot x\ast \csc x\frac{dx}{dy}=\csc x+y$
Put $\csc x=t$
$\Rightarrow -\csc x\ast \cot x\frac{dx}{dy}=\frac{dt}{dy}$
Now D.E is $\frac{dt}{dy}=-t-y$

I.F. =$e^{\int dy}=e^y$
solution is $t.e^y=\int -y{e}^{y}dy=-y{e}^y+e^y+c$
$\Rightarrow \csc x=-y+1+C{e}^{-y}$