Question

Question: (a) The differential equation in Problem 27 is equivalent to the normal form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\sqrt{\frac{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$in the square region in the-plane defined by$\mathbf{}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{<}\mathbf{1}\mathbf{,}\mathbf{\left|}\mathbf{y}\mathbf{\right|}\mathbf{<}\mathbf{1}$. But the quantity under the radical is nonnegative also in the regions defined by$\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{>}\mathbf{1}\mathbf{,}\mathbf{\left|}\mathbf{y}\mathbf{\right|}\mathbf{>}\mathbf{1}$. Sketch all regions in the-plane for which this differential equation possesses real solutions.

(b) Solve the DE in part (a) in the regions defined by.Then find an implicit and an explicit solution of the differential equation subject to$\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

Answer:

(b) The explicit solution is $y=x$.

Step-by-step solution

Sketch figure(a)

The differential equation possess solutions in the following region.

The cream region defines:$\left|x\right|<1,\left|y\right|<1$

The gray regions, which extend to infinity, define:$\left|x\right|>1,\left|y\right|>1$.

Separate variable(b)

$\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

Given the differential equation

First break up the square root and then separate the variables.

$$\begin{gathered} \frac{dy}{dx}=\sqrt{1-y^2} \\ \frac{dy}{\sqrt{1-y^2}}=\frac{dx}{\sqrt{1-x^2}} \end{gathered}$$

However the terms inside the square roots are negative for$\left|x\right|>1,\left|y\right|>1$

We need to rewrite the terms inside the radicals such that the square roots are defined.

$\frac{dy}{\sqrt{y^2-1}}=\frac{dx}{\sqrt{x^2-1}}$

However the terms inside the square roots are negative for$\left|x\right|>1,\left|y\right|>1$.

Solve integral

We need to rewrite the terms inside the radicals such that the square roots are defined.

$\frac{dy}{\sqrt{y^2-1}}=\frac{dx}{\sqrt{x^2-1}}.........\left(2\right)$

Differential equation (2) is defined for $\left|y\right|>1and\left|x\right|>1$which is equal to differential equation (1) written in normal form. To solve equation (2) we integrate the left hand side in terms of and the right hand side in terms of x.

$$\begin{gathered} \int \frac{dy}{\sqrt{y^2-1}}=\int \frac{dx}{\sqrt{x^2-1}} \\ \Rightarrow {\cosh }^{-1}y={\cosh }^{-1}x+c \end{gathered}$$

Find constant

Apply initial condition$y\left(2\right)=2$to solve for the constant.

$$\begin{gathered} {\cosh }^{-1}\left(2\right)={\cosh }^{-1}\left(2\right)+c \\ 0=c \end{gathered}$$

The implicit solution is${\cosh }^{-1}y={\cosh }^{-1}x$

Solve forto get explicit solution.

$$\begin{gathered} cos{h}^{-1}y={\cosh }^{-1}x \\ {\cosh }^{-1}\left({\cosh }^{-1}y\right)\left.={\cosh }^{-1}({\cosh }^{-1}x\right) \\ y=x \end{gathered}$$