Question

The graph of the function $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$ passing through the point $(0,1)$ and satisfying the differential equation$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}.\cos \mathrm{x}=\cos \mathrm{x}$ is such that

• A. It is a constant function
• B. It is periodic
• C. It is neither an even nor an odd function
• D. It is continuous and differentiable for all values of $\mathrm{x}$

Answer 1

Answer: (A), (B), (D)

It is continuous and differentiable for all values of $\mathrm{x}$

Given:

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycosx}=\mathrm{cosx} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{cosx}-\mathrm{ycosx} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x}(1-\mathrm{y}) \\ \frac{{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}}{1-\mathrm{y}}=\cos \mathrm{x} \end{gathered}$$

Now, integrate both sides,

$\begin{array}{rcl}\int \frac{{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}}{1-\mathrm{y}}\mathrm{dx}& =& \int \cos \mathrm{x}.\mathrm{dx}\\ -\log (1-\mathrm{y})& =& \sin \mathrm{x}+\mathrm{C}\\ \mathrm{y}& =& \left(\frac{-\mathrm{c}}{{\mathrm{e}}^{\sin \mathrm{x}}}\right)+1\end{array}$

Since, the graph of the function passes through the point $(0,1)$,

$$\begin{gathered} 1=-\mathrm{C}+1 \\ -\mathrm{C}=0 \end{gathered}$$

$\mathrm{C}=0$, which gives $\mathrm{y}=1$

.$\because \mathrm{y}=1$, it's a constant function.

Hence, the options (a), (b) and (d) are correct.