Question

find particular solution of ( y - sin x)dx + tanx dy =0

Answer 1

not realate d to my grade level

Answer 2

Answer:

$y= \frac{log secx}{sinx}+\frac{c}{sinx}$

Step-by-step explanation:

Given : $(y-sin x)dx+(tanx)dy =0$

To find : Particular solution

Solution :  $(y-sin x)dx+(tanx)dy =0$

$\rightarrow(tanx)dy =(sin x-y)dx$

$\rightarrow\frac{dy}{dx}=\frac{(sin x-y)}{tanx}$

$\rightarrow\frac{dy}{dx}=\frac{1}{cosx}-\frac{y}{tanx}$

$\rightarrow\frac{dy}{dx}+\frac{y}{tanx}=\frac{1}{cosx}$

Differential equation is of the form :

$\frac{dy}{dx}+Py=Q$

where $P=\frac{1}{tanx}$ and $Q=\frac{1}{cosx}$

Finding integrating factor $e^{\int P dx }$

IF = $e^{\int\frac{1}{tanx} dx }$

IF = $e^{\int cotx dx }$

IF = $e^{log sinx}$

IF = $sinx$

Solution → $y\times IF= \int (Q\times IF)dx +c$

∴  $y\times sinx= \int (\frac{1}{cosx}\times sinx)dx +c$

$\rightarrow y\times sinx= \int (\frac{sinx}{cosx})dx +c$

$\rightarrow y\times sinx= \int tanx dx +c$

$\rightarrow y\times sinx= log secx +c$

$\rightarrow y= \frac{log secx}{sinx}+\frac{c}{sinx}$

Therefore, the solution is $y= \frac{log secx}{sinx}+\frac{c}{sinx}$