MCQ
The solution of $\frac{{dy}}{{dx}} + 2y\,\tan x = \sin x$, is
- A$y\,{\sec ^3}x = {\sec ^2}x + c$
- ✓$y\,{\sec ^2}x = \sec x + c$
- C$y\,\,\sin x = \tan x + c$
- DNone of these
$\frac{{dy}}{{dx}} + y\,f(x) = g(x)$
$I.F.$$ = {e^{\int {f(x)dx} }} = {e^{\int {2\tan x\,dx} }} = {e^{2\log (\sec x)}} = {e^{\log {{\sec }^2}x}} = {\sec ^2}x$
Hence, the solution is $y\,({\rm{I}}{\rm{.F}}{\rm{.)}} = \int {g(x)\,{\rm{I}}{\rm{.F}}{\rm{.}}\,dx + c} $
$y({\sec ^2}x) = \int {\sin x\,{{\sec }^2}x\,dx + c} $
==> $y{\sec ^2}x = \int {\sec x\,\tan x\,dx + c} $ ==> $y{\sec ^2}x = \sec x + c$.
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