Name: Formation of Differential Equations
Uploaded: 2019-07-19
Duration: 11 min 9 s
Description: In this video, we are going to learn about the formation of differential equations in detail. Master this concept in just a few minutes with this quick lesson.

Question

Find the particular solution of the differential equation

(x - y) \frac{d y}{d x} = x + 2 y

, given that when

x = 1, y = 0 .

Accepted Answer

The given differential equation is, $(x - y) \frac{d y}{d x} = x + 2 y$

$\Rightarrow \frac{d y}{d x} = \frac{x + 2 y}{x - y}$

This is a homogeneous differential equation.

Putting $y = v x$ we get, $\frac{d y}{d x} = v + x \frac{d v}{d x}$

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

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

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

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

Integrating both sides, we get

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

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

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

$\Rightarrow \frac{3}{2} \int \frac{1}{1 + v + v^{2}} d v - \frac{1}{2} \int \frac{2 v + 1}{1 + v + v^{2}} d v = \int \frac{1}{x} d x$

$\Rightarrow \frac{3}{2} \int \frac{1}{1 + v + v^{2} + \frac{1}{4} - \frac{1}{4}} d v - \frac{1}{2} \int \frac{2 v + 1}{1 + v + v^{2}} d v = \int \frac{1}{x} d x$

$\Rightarrow \frac{3}{2} \int \frac{1}{{(v + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v - \frac{1}{2} \int \frac{2 v + 1}{1 + v + v^{2}} d v = \int \frac{1}{x} d x$

$\Rightarrow \sqrt{3} \tan^{- 1} (\frac{2 v + 1}{\sqrt{3}}) - \frac{1}{2} \log |1 + v + v^{2}| = \log | x | + C$

Putting $v = \frac{y}{x},$ we get

$\sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) - \frac{1}{2} \log |\frac{x^{2} + x y + y^{2}}{x^{2}}| = \log | x | + C$

$\Rightarrow \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) - \frac{1}{2} \log |x^{2} + x y + y^{2}| + \log | x | = \log | x | + C$

$\Rightarrow \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) - \frac{1}{2} \log |x^{2} + x y + y^{2}| = C \dots \dots$ $(1)$

Putting $x = 1$ and $y = 0$ in $(1),$ we get

$\sqrt{3} \tan^{- 1} \frac{1}{\sqrt{3}} - \frac{1}{2} \log | 1 | = C$

$\Rightarrow C = \sqrt{3} \tan^{- 1} \frac{1}{\sqrt{3}}$

$\Rightarrow C = \sqrt{3} \times \frac{π}{6}$

$\Rightarrow C = \frac{π}{2 \sqrt{3}}$

Substituting the value of $C$ in $(1),$ we get

$\Rightarrow \log |x^{2} + x y + y^{2}| = 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) - \frac{π}{\sqrt{3}}$

is the required solution.

Find the particular solution of the differential equation $(x - y) \frac{d y}{d x} = x + 2 y$ , given that when $x = 1, y = 0 .$

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Find the particular solution of the differential equation (x-y)dydx=x+2y, given that when x=1, y=0.

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Find the particular solution of the differential equation $(x - y) \frac{d y}{d x} = x + 2 y$ , given that when $x = 1, y = 0 .$