Question

solution of differential equation dy/dx=y/x+cot(y/x) is​

Answer 1

Step-by-step explanation:

We have,

$\frac{dy}{dx} = \frac{y}{x} + \cot \bigg( \frac{y}{x} \bigg)$

$\tt \large{\red{ • \: Let \: \: y = vx }} \\ \red{ \tt{\implies \: \frac{dy}{dx} = v + x \frac{dv}{dx}} }$

So,

$v + x \frac{dv}{dx} = v + \cot( v )$

$\implies \: x \frac{dv}{dx} = \cot( v )$

$\implies \: \frac{dv}{ \cot(v) } = \frac{dx}{x}$

$\implies \: \tan(v) \: dv = \frac{dx}{x}$

Integrating both sides,

$\implies \: \int \tan(v) \: dv = \int \frac{dx}{x}$

$\implies \: \sec^{2} (v) = \ln(x) + C$

$\implies \: \sec^{2} \bigg( \frac{y}{x} \bigg) = \ln(x) + C$

Answer 2

$\large\underline{\sf{Solution-}}$

The given Differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{y}{x} + cot\dfrac{y}{x}$

This is homogeneous differential equation, So solve such differential equation, we substitute,

$\boxed{ \bf{ \:y \: = \: vx \: }} - - - - (1)$

On substituting this value in above differential equation, we get

$\rm :\longmapsto\:\dfrac{d}{dx} vx= \dfrac{vx}{x} + cot\dfrac{vx}{x}$

$\rm :\longmapsto\:v\dfrac{d}{dx}x + x\dfrac{d}{dx}v = v + cotv$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = v + cotv$

$\rm :\longmapsto\:x\dfrac{dv}{dx} = cotv$

On separating the variables, we get

$\rm :\longmapsto\:\dfrac{dv}{cotv} = x \: dx$

$\rm :\longmapsto\:tanv \: dv = x \: dx$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int tanv \: dv =\displaystyle\int x \: dx$

$\rm :\longmapsto\:log |secv| = \dfrac{ {x}^{2} }{2} + c$

On substituting the value of v, we get

$\rm :\longmapsto\:log \bigg|sec \dfrac{y}{x} \bigg| = \dfrac{ {x}^{2} }{2} + c$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$