Differential Equations — Class 12 Maths Solution

: The given equation is $\cfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = - 3\cot x$ and $Q = \sin 2x$

$\therefore$ $\int P dx = - 3\int {\cot } xdx = - 3\log |\sin x|$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{ - 3\log \left| {\sin x} \right|}} = {e^{\log {\rm{cose}}{{\rm{c}}^3}x}} = {\rm{cose}}{{\rm{c}}^3}x$

$\therefore$ The solution is $y \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right) = \int Q ({\rm{I}}.{\rm{F}}.)dx + C$

$\Rightarrow y\,{\rm{cose}}{{\rm{c}}^3}x = \int {\sin } 2x\,{\rm{cose}}{{\rm{c}}^3}xdx + C = \int {\cfrac{{2\sin x\cos x}}{{{{\sin }^3}x}}} dx + C$

$= 2\int {\cos {\rm{ec}}x\cot xdx} + C = - 2\cos {\rm{ec}}x + C$

$\Rightarrow y = - 2{\rm{si}}{{\rm{n}}^2}x + C{\sin ^3}x$

…(2)
When $x = \cfrac{\pi }{2},y = 2\,\,\,\,\therefore \,\,\,\,2 = - 2{\sin ^2}\cfrac{\pi }{2} + C{\sin ^3}\cfrac{\pi }{2}$

$\Rightarrow 2 = - 2{(1)^2} + C{(1)^3} \Rightarrow C = 2 + 2 = 4$

Putting $C = 4$ in (2),

we get
$y = - 2{\sin ^2}x + 4{\sin ^3}x\; \Rightarrow y = 4{\sin ^3}x - 2{\sin ^2}x$

which is the required solution.