For x ∈ ℝ , let the function y(x) be the solution of the...

Given,

$\dfrac{dy}{dx} + 12y = \cos \left(\dfrac{\pix}{12}\right)$

$I . F = {e}^{12x}$

Now solution of the differential equation is given by,

$⇒ y \cdot {e}^{12x} = ∫ {e}^{12x} \cdot \cos \left(\dfrac{\pix}{12}\right)dx + C$

$⇒ y \cdot {e}^{12x} = \dfrac{{e}^{12x}}{{12}^{2} + {\left(\dfrac{\pi}{12}\right)}^{2}}\left[12cos\dfrac{\pix}{12} + \dfrac{\pi}{12}\sin \left(\dfrac{\pix}{12}\right)\right] + C$

Now given curve is passing through \left(0, 0\right)$</p><p>So,$y\left(0\right) = 0 ⇒ C = - \dfrac{12}{{12}^{2} + {\left(\dfrac{\pi}{12}\right)}^{2}}$</p><p>So$y = \dfrac{1}{\lambda}\left[\mathop{\mathop{12\cos \left(\dfrac{\pix}{12}\right) + \dfrac{\pi}{12}\sin \left(\dfrac{\pix}{12}\right)}_{⏟}}_{{f}_{1}\left(x\right)} - 12{e}^{- 12x}\right]$</p><p>{Where \lambda = \dfrac{1}{{12}^{2} + {\left(\dfrac{\pi}{12}\right)}^{2}} }$

Now on differentiating we get,

$\dfrac{dy}{dx} = \dfrac{1}{{\lambda}^{'}}\left[\mathop{\mathop{- \pi sin\left(\dfrac{\pix}{12}\right) + \dfrac{{\pi}^{2}}{{12}^{2}}\cos \left(\dfrac{\pix}{12}\right)}_{⏟}}_{{f}_{2}\left(x\right)} + 12{e}^{- 12x}\right]$

\left\{\text{where}{\lambda}^{'} = \dfrac{12}{{12}^{2} + {\left(\dfrac{\pi}{12}\right)}^{2}}\right\}$</p><p>Now when$x$is large then$12{e}^{- 12x}$tends to zero.</p><p>But${f}_{2}\left(x\right)$varies in$\left[- \sqrt{{\pi}^{2} + {\left(\dfrac{\pi}{12}\right)}^{4}}, \sqrt{{\pi}^{2} + {\left(\dfrac{\pi}{12}\right)}^{4}}\right] \left\{\text{by using} - \sqrt{{a}^{2} + {b}^{2}} \text{ and } \sqrt{{a}^{2} + {b}^{2}}\right\}$</p><p>Hence$\dfrac{dy}{dx}$is changing its sign.</p><p>So$y\left(x\right)$is non monotonic for all real number.<br />Also when$x$is very large then again$- 12{e}^{- 12x} is almost zero but ${f}_{1}$\left(x\right)$is periodic, so there exist some$\beta$for which$y = \beta$intersect$y = y$\left(x\right)$ at infinitely many points.