Question

Find c if dy/dx — y tan x = 3 e^—sin x if y = 4 and x= 0

Answer 1

Answer:

1. I looking my text book I explanation

Step-by-step explanation:

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Answer 2

The value of constant 'c' for the given differential equation is 1.

Step-by-step explanation:

Given: Differential equation $\dfrac{dy}{dx}-y tan x = 3 e^{-sin x}$

To Find: The value of constant 'c' for y = 4 and x = 0

Solution:

• Finding the value of constant 'c' for the given differential equation when y = 4 and x = 0

The given differential equation has the form $\frac{dy}{dx} +Py = Q$ such that we have $P = - tanx$ and $Q = 3e^{-sinx}$. Therefore, It is a linear differential equation.

The Integrating Factor (I.F.) is $\Rightarrow I.F. = e^{\int{Pdx}}$

$\Rightarrow I.F. = e^{\int{(-tanx)dx}} = e^{ln(secx)dx}} = secx$

The solution of the differential equation is given by,

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

$\Rightarrow y (secx) = 3 \int (e^{-sinx}secx) dx + c \ \cdots \cdots (2)$

Now, considering $I = \int (e^{-sinx}secx)$ such that its integral can be calculated by the method of 'integration by parts'

$I = secx \int e^{-sinx} dx - \dfrac{d(secx)}{dx} \int \int (e^{-sinx})dx\\\\I = secx (e^{-sinx} secx) - secx \ tanx \int (e^{-sinx} secx)dx\\\\I = secx (e^{-sinx} secx) - secx \ tanx ( I)$

$I (1 + secx \ tanx ) = secx (e^{-sinx} secx)\\\\I = \dfrac{e^{-sinx} sec^{2} x}{1 + secx \ tanx} \ \cdots \cdots (2)$

Now, substituting (2) in (1), we get,

$\Rightarrow y (secx) = \dfrac{3e^{-sinx} sec^{2} x}{1 + secx \ tanx} + c$

$\Rightarrow y = \dfrac{3e^{-sinx} secx}{1 + secx \ tanx} + \dfrac{c}{secx}$

Now, using the conditions x = 0 and y = 4 in the above solution, we get,

$\Rightarrow 4 = \dfrac{3e^{-sin(0)} sec(0)}{1 + sec(0) \ tan(0)} + \dfrac{c}{sec(0)}$

$\Rightarrow c + 3 = 4 \Rightarrow c = 4 - 3 = 1$

Hence, the value of constant 'c' is 1 for the given differential equation when y = 4 and x = 0.