2.60   ODE No. 60

1. Problem in Latex
2. Mathematica input
3. Maple input

$$y^{\prime }(x)-\frac{\sqrt{y(x{)}^2-1}}{\sqrt{x^2-1}}=0$$ ✓ Mathematica : cpu = 0.111424 (sec), leaf count = 173

$$\{\{y(x)\to -\frac{1}{2}{e}^{-c_1}\sqrt{2x^2+2e^{4c_1}{x}^2-2\sqrt{(x-1)(x+1)}x+2e^{4c_1}\sqrt{(x-1)(x+1)}x-1+2e^{2c_1}-e^{4c_1}}\},\{y(x)\to \frac{1}{2}{e}^{-c_1}\sqrt{2x^2+2e^{4c_1}{x}^2-2\sqrt{(x-1)(x+1)}x+2e^{4c_1}\sqrt{(x-1)(x+1)}x-1+2e^{2c_1}-e^{4c_1}}\}\}$$ ✓ Maple : cpu = 0.016 (sec), leaf count = 29

$$\{\ln (x+\sqrt{x^2-1})-\ln (y\left(x\right)+\sqrt{{\left(y\left(x\right)\right)}^2-1})+\mathit{\_}\mathit{C}\mathit{1}=0\}$$

Hand solution

$$\begin{array}{}\text{(1)}& y^{\prime }=\pm \frac{\sqrt{y^2-1}}{\sqrt{x^2-1}}\end{array}$$

Separable. For the positive case

$$\begin{array}{rl}\frac{dy}{dx}\frac{1}{\sqrt{y^2-1}}& =\frac{1}{\sqrt{x^2-1}}\\ \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}& =\frac{dx}{{(x^2-1)}^{\frac{1}{2}}}\end{array}$$

Integrating

$$\int \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}=\int \frac{dx}{{(x^2-1)}^{\frac{1}{2}}}+C$$

But $\int \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}={\tanh }^{-1}\frac{y}{{(y^2-1)}^{\frac{1}{2}}}=\ln (y+\sqrt{y^2-1})$, hence

$$\ln (y+\sqrt{y^2-1})=\ln (x+\sqrt{x^2-1})+C$$

For the negative case

$$\begin{array}{rl}\frac{dy}{dx}\frac{1}{\sqrt{y^2-1}}& =-\frac{1}{\sqrt{x^2-1}}\\ \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}& =-\frac{dx}{{(x^2-1)}^{\frac{1}{2}}}\end{array}$$

Integrating

$$\int \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}=-\int \frac{dx}{{(x^2-1)}^{\frac{1}{2}}}+C$$

But $\int \frac{dy}{{(y^2-1)}^{\frac{1}{2}}}={\tanh }^{-1}\frac{y}{{(y^2-1)}^{\frac{1}{2}}}=\ln (y+\sqrt{y^2-1})$, hence

$$\ln (y+\sqrt{y^2-1})=-\ln (x+\sqrt{x^2-1})+C$$

Therefore

$$\ln (y+\sqrt{y^2-1})=\pm \ln (x+\sqrt{x^2-1})+C$$