Q10RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Decide whether the statement made is True or False. The relation $\mathbf{sin} \mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}\mathbf{=}{\mathbf{x}}^{\mathbf{6}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}$  is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{6}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{cos} \mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}}$.

Step-by-Step Solution

Answer

The statement is true.

Using the differential formula

For the result use the differential formula  $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{(}{\mathbf{x}}^{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{n} {\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$ and consider x and y as variable.

Differentiating sin   y + e y = x 6 - x 2 + x + 1 with respect to x.

The Solution is given below,

$$\begin{gathered} \frac{\mathbf{d}}{\mathbf{dx}}\left(\mathbf{sin} \mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}\right)\mathbf{=}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{(}{\mathbf{x}}^{\mathbf{6}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{cosy}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1} \\ \mathbf{(}\mathbf{cosy}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}\mathbf{)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1} \\ \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{6}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{cosy}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}} \end{gathered}$$

Hence, this is the given differential equation, the given statement is true.