Answer to Question #106446 in Mechanics | Relativity for Najeya

Question

Solve dy/dt + ycotx = e^cosx for x= pi by two and y= -2

Answer 1

Mechanics | Relativity

We need to solve the given Differential equation

Solution:

Given equation is the Linear differential in the form $\frac {dy}{dx} + P(x) y = Q(x)$

Now compare the given equation $\frac {dy}{dx} + y \space cot x = e^ {cos \space x}$ with the above form

$$P(x) = cot \space x$$

$$Q(x) = e^{cos \space x}$$

Integrating factor = I.F = $e^{\int P(x) dx} = e^ {\int cot x \space dx} = e^ {ln \space sin x} = sin \space x$

The solution of the Linear equation is,

$$y \times (I.F) = \int (I.F) \times Q(x) dx + c$$

$$y \times sin x = \int sin x \space e^{cos x} dx + c$$

$$y \space sin x = - \int d(cos x) \space e^{cos x} dx + c$$

$$y \space sin x = - e^{cos x} + c$$

Now we need to find the c value by plugging $x = \frac {\pi}{2} \space and \space y =-2$

$$-2 (sin \frac{\pi}{2} ) = - e^{cos {\frac {\pi}{2 }}} + c$$

$$-2 =- e^0 + c \\ -2 = -1 +c \\ c = -1$$

Finally the solution is ,

$$y \space sin x = - e^{cos x} -1$$