Q12 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\left(x+y\right)\mathbf{=}\mathbf{1}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{xsec}\left(x+y\right)\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The given relation is an implicit solution to the given differential equation.

Step1: Differentiating the given relation

As, in the given relation ${\mathrm{x}}^2-\sin (\mathrm{x}+\mathrm{y})=1$, y is defined implicitly as the function of x, so by using implicit differentiation, we will differentiate the given relation concerning x,

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2-\sin \left(\mathrm{x}+\mathrm{y}\right)\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(1\right) \\ 2\mathrm{x}-\cos \left(\mathrm{x}+\mathrm{y}\right)\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \end{gathered}$$

Step 2: Simplification of the differential equation obtained in step1

$$\begin{gathered} 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\mathrm{x}}{-\cos \left(\mathrm{x}+\mathrm{y}\right)} \\ 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{x}}{\cos \left(\mathrm{x}+\mathrm{y}\right)} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\times \sec \left(\mathrm{x}+\mathrm{y}\right)-1 \end{gathered}$$

Which is identical to the given differential equation.

 

Thus, the relation ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\left(x+y\right)\mathbf{=}\mathbf{1}\mathbf{,}$ is an implicit solution to the differential equation $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{xsec}\left(x+y\right)\mathbf{-}\mathbf{1}$.