Question

Solve the initial-value problem\({y^\prime } = \frac{{(\sin x)}}{{\sin y}},l = \pi /2\), and graph the solution (if your CAS does implicit plots).

Short answer

\(\cos y = \cos x - 1\)

Step-by-step solution

Definition

Separable differential equations can be written in the form\(\frac{{dy}}{{dx}} = {\rm{ }}f\left( x \right){\rm{ }}g\left( y \right)\), where\(x,y\)are the variables and are explicitly separated from each other.

Separate variable

Consider the initial-value problem\({y^\prime } = \frac{{\sin x}}{{\sin y}}\)with the initial condition is\(y(0) = \frac{\pi }{2}\).

We know that\({y^\prime } = \frac{{dy}}{{dx}}\), and then the function as:

\(\begin{aligned}{c}{y^\prime } = \frac{{\sin x}}{{\sin y}}\\\frac{{dy}}{{dx}} = \frac{{\sin x}}{{\sin y}}\\\sin ydy = \sin xdx\end{aligned}\)

Integrate

Integrating both sides we have:

\(\int {\sin } ydy = \int {\sin } xdx\)

\( - \cos y = - \cos x - C\) Using formula\(\int {\sin } xdx = - \cos x + C\)

\(\cos y = \cos x + C\;\;\)Where\(C\)is constant of integration.

Apply initial condition

Using initial condition\(y(0) = \frac{\pi }{2}\;\)that means\(x = 0,y = \frac{\pi }{2}\).

\(\begin{aligned}{c}\cos \left( {\frac{\pi }{2}} \right) = \cos (0) + C\\0 = 1 + C\\C = - 1\end{aligned}\)

Find solution

Substitute value\(C = - 1\) in\(\cos y = \cos x + C\;\;\)we have

\(\cos y = \cos x - 1\)

Therefore, the solution of given the initial-value problem is\(\cos y = \cos x - 1\).