Question

dy/dx+y tanx=y sec x​

Answer 1

Answer:

$\large\bold\red{c = \frac{y}{|1 + \sin(x)| }}$

Step-by-step explanation:

Given,

A differential equation,

• $\frac{dy}{dx} + y \tan(x) = y \sec(x)$

To solve the given Equation,

Divide both sides with y,

We get,

$= > \frac{dy}{ydx} + \frac{y \tan(x) }{y} = \frac{y \sec(x) }{y} \\ \\ = > \frac{dy}{ydx} + \tan(x) = \sec(x) \\ \\ = > \frac{dy}{ydx} = \sec(x) - \tan(x) \\ \\ = > \frac{dy}{y} = ( \sec(x) - \tan(x) )dx$

Now,

Integrating on both sides,

We get,

$= > \int \frac{dy}{y} = \int( \sec(x) - \tan(x)) dx \\ \\ = > \int \frac{dy}{y} = \int \sec(x) dx - \int \tan(x) dx$

But,

We know that,

• $\bold{\int \frac{dy}{y} = ln(y)}$

• $\bold{\int \sec(x) dx = ln| \sec(x) + \tan(x) | }$

• $\bold{\int \tan(x) dx = - ln | \cos(x) |}$

Therefore,

Putting the respective values,

We get,

$= > ln(y) = ln | \sec(x) + \tan(x) | - ( - ln | \cos(x) |) + ln(c) \\ \\ = > ln(y) = ln | \sec(x) + \tan(x) | + ln | \cos(x) | + ln(c)$

But,

We know that,

• $\bold{ln(a) + ln(b) = ln(ab) }$

Therefore,

We get,

$= > ln(y) = ln | \cos x ( \sec x + \tan x) | + ln(c) \\ \\ = > ln(y) = ln|1 + \sin(x) | + ln(c) \\ \\ = > ln(y) = ln |c(1 + \sin x) | \\ \\ = > y = c(|1 + \sin x|) \\ \\ = > c = \frac{y}{|1 + \sin(x)| }$

Hence,

Solution of the differential equation is

• $\large\bold{c = \frac{y}{|1 + \sin(x) |} }$