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