Name: Linear Differential Equation: Illustration
Uploaded: 2019-07-19
Duration: 3 min 43 s
Description: You might be familiar with linear equations right from your middle school. Let's upgrade the concept to linear differential equations. This video teaches you...

Question

Solve the following differential equation:

$(x - y) d y - (x + y) d x = 0$

Accepted Answer

Given differential equation is $(x - y) d y - (x + y) d x = 0$

$\Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y}$

$\Rightarrow \frac{d y}{d x} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} \to (1)$

This is a homogeneous differential equation because $\frac{d y}{d x}$ is a function of $(\frac{y}{x})$ .

Let $y = v x \to (2)$

Differentiate with respect to $x$ , then $\frac{d y}{d x} = v + x \frac{d v}{d x} \to (3)$

Using $(2)$ and $(3)$ in $(1)$ , we get: $v + x \frac{d v}{d x} = \frac{1 + v}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^{2}}{1 - v}$

$\Rightarrow \frac{(1 - v) d v}{1 + v^{2}} = \frac{d x}{x}$

Integrating both sides, we get: $\int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{1 + v^{2}} d v - \int \frac{v}{1 + v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{1 + v^{2}} d v - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \log |1 + v^{2}| = \log |x| + c$ [Using $\int \frac{f^{'} (x)}{f (x)} d x = \log |f (x)| + c$ ]

$\Rightarrow \tan^{- 1} (\frac{y}{x}) = \frac{1}{2} \log |1 + {(\frac{y}{x})}^{2}| + \log |x| + c$

$\Rightarrow 2 \tan^{- 1} (\frac{y}{x}) = \log |1 + {(\frac{y}{x})}^{2}| + 2 \log |x| + 2 c$

$\Rightarrow 2 \tan^{- 1} (\frac{y}{x}) = \log (1 + {(\frac{y}{x})}^{2}) + \log x^{2} + 2 c$

$\Rightarrow 2 \tan^{- 1} (\frac{y}{x}) = \log ((1 + \frac{y^{2}}{x^{2}}) \times x^{2}) + 2 c$

$\Rightarrow 2 \tan^{- 1} (\frac{y}{x}) = \log (x^{2} + y^{2}) + c_{1}, c_{1} = 2 c$

Hence, the required general solution is $2 \tan^{- 1} (\frac{y}{x}) = \log (x^{2} + y^{2}) + c_{1}$ .

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