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Chapter 2: Problem 46

(b) For \(|x|>1\) and \(|y|>1\) the differential equation is \(d y / d x=\sqrt{y^{2}-1} / \sqrt{x^{2}-1} .\) Separating variables and integrating, we obtain \(\frac{d y}{\sqrt{y^{2}-1}}=\frac{d x}{\sqrt{x^{2}-1}} \quad\) and \(\quad \cosh ^{-1} y=\cosh ^{-1} x+c\). Setting \(x=2\) and \(y=2\) we find \(c=\cosh ^{-1} 2-\cosh ^{-1} 2=0\) and \(\cosh ^{-1} y=\cosh ^{-1} x .\) An explicit solution is \(y=x\).

Short Answer

Expert verified
The explicit solution is \( y = x \).

Step by step solution

01

State the Problem

We need to solve the differential equation \( \frac{dy}{dx} = \frac{\sqrt{y^2 - 1}}{\sqrt{x^2 - 1}} \) given the conditions \( |x| > 1 \) and \( |y| > 1 \).
02

Separate the Variables

Rewrite the differential equation by separating the variables, giving us \( \frac{dy}{\sqrt{y^2 - 1}} = \frac{dx}{\sqrt{x^2 - 1}} \).
03

Integrate Both Sides

Integrate both sides of the equation separately: \[ \int \frac{dy}{\sqrt{y^2 - 1}} = \int \frac{dx}{\sqrt{x^2 - 1}} \] The integral of \( \frac{1}{\sqrt{u^2 - 1}} \) is \( \cosh^{-1} u \), so we obtain\[ \cosh^{-1} y = \cosh^{-1} x + C \] where \( C \) is a constant of integration.
04

Substitute Initial Conditions

We are given that when \( x = 2 \) and \( y = 2 \), the relationship holds. Substitute these values into the equation:\[ \cosh^{-1} 2 = \cosh^{-1} 2 + C \] This simplifies to \( C = 0 \).
05

Solve for y in terms of x

With the constant \( C = 0 \), the equation becomes \( \cosh^{-1} y = \cosh^{-1} x \). Since the inverse hyperbolic cosine is a one-to-one function, it implies \( y = x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
When faced with a differential equation, one effective method to solve it is through separation of variables. This approach involves rearranging the equation so that each variable appears on a different side of the equation. It is essentially a way to "separate" the variables involved. In our example, the original differential equation is given as: \( \frac{dy}{dx} = \frac{\sqrt{y^2 - 1}}{\sqrt{x^2 - 1}} \). To separate the variables, we rearrange it to: \( \frac{dy}{\sqrt{y^2 - 1}} = \frac{dx}{\sqrt{x^2 - 1}} \). By doing so, each side of the equation only involves one variable, allowing us to integrate each side separately. This step is crucial because it lays the foundation for finding a solution to the differential equation.
Hyperbolic Functions
Hyperbolic functions are similar to trigonometric functions but are expressed through hyperbolas instead of circles. The most common hyperbolic functions include the hyperbolic sine, cosine, and tangent, represented as \( \sinh \), \( \cosh \), and \( \tanh \) respectively. In our problem, we specifically encounter \( \cosh^{-1} \), which is the inverse hyperbolic cosine function. This function is vital in reducing integrals that involve forms of \( \frac{1}{\sqrt{u^2 - 1}} \) as in our differential equation.The integration leads us to \( \cosh^{-1} y = \cosh^{-1} x + C \). Hyperbolic functions have properties similar to their trigonometric counterparts. Knowing these can help simplify and solve various equations involving hyperbolas.
Initial Conditions
In mathematics, initial conditions are used to find particular solutions to differential equations. These conditions provide specific values for the variables involved, allowing us to determine any unknown constants in the equation.In our example, the initial conditions are given as \( x = 2 \) and \( y = 2 \). When we substitute these into the integrated equation: \( \cosh^{-1} 2 = \cosh^{-1} 2 + C \),we find that the constant \( C \) equals 0. Having initial conditions turns the general solution of a differential equation into a unique solution that satisfies the specific conditions at a given point.
Integration of Rational Functions
Integration of rational functions involves integrating expressions that are ratios of polynomials. It is often necessary to use substitution or other integration techniques to simplify these expressions.In our exercise, both sides of the separated differential equation: \( \int \frac{dy}{\sqrt{y^2 - 1}} \) and \( \int \frac{dx}{\sqrt{x^2 - 1}} \)are examples of rational functions after being transformed through separation. By recognizing that these integrals are forms of inverse hyperbolic functions, we can easily integrate them to achieve \( \cosh^{-1} \).Mastering the integration of rational functions is key to solving a broad range of differential equations using the method of separation of variables.

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Most popular questions from this chapter

From \(\frac{y}{\left(1+y^{2}\right)^{1 / 2}} d y=\frac{x}{\left(1+x^{2}\right)^{1 / 2}} d x\) we obtain \(\left(1+y^{2}\right)^{1 / 2}=\left(1+x^{2}\right)^{1 / 2}+c\).

The linear equation \(d x / d t=-\lambda_{1} x\) can be solved by either separation of variables or by an integrating factor. Integrating both sides of \(d x / x=-\lambda_{1} d t\) we obtain \(\ln |x|=-\lambda_{1} t+c\) from which we get \(x=c_{1} e^{-\lambda_{1} t} .\) Using \(x(0)=x_{0}\) we find \(c_{1}=x_{0}\) so that \(x=x_{0} e^{-\lambda_{1} t} .\) Substituting this result into the second differential equation we have $$\frac{d y}{d t}+\lambda_{2} y=\lambda_{1} x_{0} e^{-\lambda_{1} t}$$ which is linear. An integrating factor is \(e^{\lambda_{2} t}\) so that $$\frac{d}{d t}\left[e^{\lambda_{2} t} y\right]=\lambda_{1} x_{0} e^{\left(\lambda_{2}-\lambda_{1}\right) t}+c_{2}$$ $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{\left(\lambda_{2}-\lambda_{1}\right) t} e^{-\lambda_{2} t}+c_{2} e^{-\lambda_{2} t}=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+c_{2} e^{-\lambda_{2} t}$$ Using \(y(0)=0\) we find \(c_{2}=-\lambda_{1} x_{0} /\left(\lambda_{2}-\lambda_{1}\right) .\) Thus $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Substituting this result into the third differential equation we have $$\frac{d z}{d t}=\frac{\lambda_{1} \lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Integrating we find $$z=-\frac{\lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}+c_{3}$$ Using \(z(0)=0\) we find \(c_{3}=x_{0} .\) Thus $$z=x_{0}\left(1-\frac{\lambda_{2}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}\right)$$.

For \(\frac{d T}{d t}-k T=-T_{m} k\) an integrating factor is \(e^{\int(-k) d t}=e^{-k t}\) so that \(\frac{d}{d t}\left[e^{-k t} T\right]=-T_{m} k e^{-k t}\) and \(T=T_{m}+c e^{k t}\) for \(-\infty< t<\infty .\) If \(T(0)=T_{0}\) then \(c=T_{0}-T_{m}\) and \(T=T_{m}+\left(T_{0}-T_{m}\right) e^{k t}\).

For \(y^{\prime}+\frac{2}{x^{2}-1} y=\frac{x+1}{x-1}\) an integrating factor is \(e^{\int\left[2 /\left(x^{2}-1\right)\right] d x}=\frac{x-1}{x+1}\) so that \(\frac{d}{d x}\left[\frac{x-1}{x+1} y\right]=1\) and \((x-1) y=x(x+1)+c(x+1)\) for \(-1< x<1\).

(a) After separating variables we obtain \\[ \begin{aligned} \frac{m d v}{m g-k v^{2}} &=d t \\ \frac{1}{g} \frac{d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \frac{\sqrt{m g}}{\sqrt{k} g} \frac{\sqrt{k / m g} d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \sqrt{\frac{m}{k g}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=t+c \\ \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=\sqrt{\frac{k g}{m}} t+c_{1} \end{aligned} \\] Thus the velocity at time \(t\) is \\[ v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) \\] Setting \(t=0\) and \(v=v_{0}\) we find \(c_{1}=\tanh ^{-1}\left(\sqrt{k} v_{0} / \sqrt{m g}\right).\) (b) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) we have \(v \rightarrow \sqrt{m g / k}\) as \(t \rightarrow \infty\) (c) Integrating the expression for \(v(t)\) in part (a) we obtain an integral of the form \(\int d u / u:\) \\[ s(t)=\sqrt{\frac{m g}{k}} \int \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) d t=\frac{m}{k} \ln \left[\cosh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right)\right]+c_{2} \\] Setting \(t=0\) and \(s=0\) we find \(c_{2}=-(m / k) \ln \left(\cosh c_{1}\right),\) where \(c_{1}\) is given in part \((\mathrm{a}).\)
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