Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
✓
Answer
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\sqrt{\text{y}^2-\text{x}^2}+\text{y}}{\text{x}}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\sqrt{\text{v}^2\text{x}^2-\text{x}^2}+\text{vx}}{\text{x}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}$
$\Rightarrow\ \frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\text{v}+\sqrt{\text{v}^2-1}\Big|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \text{v}+\sqrt{\text{v}^2-1}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \frac{\text{y}}{\text{x}}+\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}=\text{Cx}^2$
$\Rightarrow\ \text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$
Hence, $\text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$ is the required solution.
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\sqrt{\text{y}^2-\text{x}^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\sqrt{\text{y}^2-\text{x}^2}+\text{y}}{\text{x}}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\sqrt{\text{v}^2\text{x}^2-\text{x}^2}+\text{vx}}{\text{x}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}+\text{v}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=2\sqrt{\text{v}^2-1}$
$\Rightarrow\ \frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{2\sqrt{\text{v}^2-1}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\text{v}+\sqrt{\text{v}^2-1}\Big|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \text{v}+\sqrt{\text{v}^2-1}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \frac{\text{y}}{\text{x}}+\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}=\text{Cx}^2$
$\Rightarrow\ \text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$
Hence, $\text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{Cx}^3$ is the required solution.
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