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

Solve:

y \cos \frac{y}{x} (x d y - y d x) + x \sin \frac{y}{x} (x d y + y d x) = 0

Accepted Answer

$y \cos \frac{y}{x} (x d y - y d x) + x \sin \frac{y}{x} (x d y + y d x) = 0$

$x y \cos \frac{y}{x} d y - y^{2} \cos \frac{y}{x} d x + x^{2} \sin \frac{y}{x} d y + x y \sin \frac{y}{x} d x = 0$

Dividing by $x^{2}$ , we get

$(\frac{y}{x} \cos \frac{y}{x} + \sin \frac{y}{x}) d y + (\frac{y}{x} \sin \frac{y}{x} - \frac{y^{2}}{x^{2}} \cos \frac{y}{x}) d x = 0$

$\Rightarrow \frac{d y}{d x} = \frac{{(\frac{y}{x})}^{2} \cos (\frac{y}{x}) - \frac{y}{x} \sin \frac{y}{x}}{\frac{y}{x} \cos \frac{y}{x} + \sin \frac{y}{x}} . . . (i)$

Put $\frac{y}{x} = t \Rightarrow \frac{d y}{d x} = t + x \frac{d t}{d x} . . . (ii)$

Using equations $(i)$ and $(ii)$ , we have

$t + x \frac{d t}{d x} = \frac{t^{2} \cos t - t \sin t}{t \cos t + \sin t}$

$x \frac{d t}{d x} = \frac{t^{2} \cos t - t \sin t - t^{2} \cos t - t \sin t}{t \cos t + \sin t}$

$x \frac{d t}{d x} = \frac{- 2 t \sin t}{t \cos t + \sin t}$

$\Rightarrow \int \frac{(t \cos t + \sin t)}{t \sin t} d t = - 2 \int \frac{d x}{x}$

$\Rightarrow \int \cot t d t + \int \frac{1}{t} d t = - 2 \log x + \log C$

$\Rightarrow logsin t + \log t = \log \frac{C}{x^{2}}$

$\Rightarrow t \sin t = \frac{C}{x^{2}}$

$\frac{y}{x} \sin \frac{y}{x} = \frac{C}{x^{2}} \Rightarrow x y \sin \frac{y}{x} = C$

Solve: $y \cos \frac{y}{x} (x d y - y d x) + x \sin \frac{y}{x} (x d y + y d x) = 0$

Important Questions on Differential Equations - II

Learn and Prepare for any exam you want !

Solve: ycosyx(xdy-ydx)+xsinyx(xdy+ydx)=0

Important Questions on Differential Equations - II

Learn and Prepare for any exam you want !

Solve: $y \cos \frac{y}{x} (x d y - y d x) + x \sin \frac{y}{x} (x d y + y d x) = 0$