Question
Solve the following differential equations:
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
✓
Answer
We have,
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Integrating both sides, we get
$\int\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Putting $1 + y^2= t^2$ and $1 + x^2 = u^2$, we get
$2y\ dy = 2t\ dt$ and $2x\ dx = 2u\ du$
$\Rightarrow\text{dy}=\frac{\text{t}}{\text{y}}\ \text{dt}\ \text{and}\ \text{dx}=\frac{\text{u}}{\text{x}}\ \text{du}$
$\therefore\int\frac{\text{t}^2}{\text{y}^2}\ \text{dt}=-\int\frac{\text{u}^2}{\text{x}^2}\ \text{dx}$
$\Rightarrow\int\frac{\text{t}^2}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\int\frac{\text{t}^2-1+1}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2-1+1}{\text{u}^2-1}\ \text{du}$
$\int\text{dt}+\int\frac{1}{\text{t}^2-1}\ \text{dt}=-\int\text{du}-\int\frac{1}{\text{u}^2-1}\ \text{du}$
Substituting t by $\sqrt{1+\text{y}^2}$ and u by $\sqrt{1+\text{x}^2}$
$\sqrt{1+\text{y}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=-\sqrt{1+\text{x}^2}\\-\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$
Hence, $\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$ is the required solution.
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=-\text{y}\sqrt{1+\text{x}^2}$
$\Rightarrow\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Integrating both sides, we get
$\int\frac{\sqrt{1+\text{y}^2}}{\text{y}}\ \text{dy}=-\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\ \text{dx}$
Putting $1 + y^2= t^2$ and $1 + x^2 = u^2$, we get
$2y\ dy = 2t\ dt$ and $2x\ dx = 2u\ du$
$\Rightarrow\text{dy}=\frac{\text{t}}{\text{y}}\ \text{dt}\ \text{and}\ \text{dx}=\frac{\text{u}}{\text{x}}\ \text{du}$
$\therefore\int\frac{\text{t}^2}{\text{y}^2}\ \text{dt}=-\int\frac{\text{u}^2}{\text{x}^2}\ \text{dx}$
$\Rightarrow\int\frac{\text{t}^2}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\int\frac{\text{t}^2-1+1}{\text{t}^2-1}\ \text{dt}=-\int\frac{\text{u}^2-1+1}{\text{u}^2-1}\ \text{du}$
$\int\text{dt}+\int\frac{1}{\text{t}^2-1}\ \text{dt}=-\int\text{du}-\int\frac{1}{\text{u}^2-1}\ \text{du}$
Substituting t by $\sqrt{1+\text{y}^2}$ and u by $\sqrt{1+\text{x}^2}$
$\sqrt{1+\text{y}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=-\sqrt{1+\text{x}^2}\\-\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$
Hence, $\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{y}^2}-1}{\sqrt{1+\text{y}^2}+1}\Big|=\text{C}$ is the required solution.
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