Question

For each of the differential equation, find a particular solution satisfying the given condition: $\frac{dy}{dx}$ =y tan x ; y = 1 when x = 0

Answer 1

Answer:

y = sec(x)

Step-by-step explanation:

Hi,

Given that dy/dx = ytanx

We can write dy/y = tanx dx

Integrating on both sides, we get

$\int\ {dy/y} \, dx$ = $\int\ {tanx} \, dx$

㏑|y| = ln|sec x| + c, where c is an arbitrary constant

or we can write

㏑|y/sec x| = c

y/sec x = ${e^{c}}$ = k(say)

y = ksec x,----------------(1)

Also, given that y = 1 , when x = 0

So,  substituting in (1), we get

1 = k*sec(0) = k

Hence, value of k = 1.

So, equation (1), will become y = sec(x).

Hence, y = sec(x) is the required particular solutions.

Hope, it helps !