Question

How do you solve the initial-value problem $y' = \dfrac{{\sin x}}{{\sin y}}$ where $y(0) = \dfrac{\pi }{4}$ ?

Answer 1

Hint: To solve the initial value problem, we have to integrate the given function to get a function in variable $y$ and $x$ . And then apply the given values of variable $x$ and variable $y$ to get a particular solution of the given function.

Complete step-by-step solution:
Here, we are given a derivative of the function $y(x)$ .
Hence, we need to apply integration to the given function to get a function in variable $x$ and $y$ .
Here, we are given the derivative as $\;y'$ . It can also be written as
$y'(x) = \dfrac{{dy}}{{dx}}$
Substituting in the given equation, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{\sin x}}{{\sin y}}$
Multiplying $\;dx$ on both sides of the equation,
$\dfrac{{dy}}{{dx}} \times dx = \dfrac{{\sin x}}{{\sin y}} \times dx$
$\Rightarrow dy = \dfrac{{\sin x}}{{\sin y}} \times dx$
Multiplying $\sin x$ on both sides of the equation
$\Rightarrow dy \times \sin y = \dfrac{{\sin x}}{{\sin y}} \times dx \times \sin y$
$\Rightarrow dy \times \sin y = \sin x \times dx$
Applying integration on both sides of the equation
$\Rightarrow \int {\sin ydy} = \int {\sin xdx}$
We know that the integration of $\sin x$ is shown as
$\Rightarrow \int {\sin xdx} = - \cos x + c$
Applying the solution on both sides of the equation,
$\Rightarrow - \cos y = - \cos x + c$
Multiplying $( - 1)$ on both side of the equation
$\Rightarrow ( - 1) \times - \cos y = ( - 1) \times \cos x + ( - 1) \times c$
$\Rightarrow \cos y = \cos x - c$
This can be called the general solution of the given derivation
Now, we are given the initial value of this problem as
$y(0) = \dfrac{\pi }{4}$
This can be explained as when the value of $x$ is taken as $x = 0$ the value of the variable $y$ will be $y = \dfrac{\pi }{4}$
Now, substituting these values in the general solution
$\Rightarrow \cos \left( {\dfrac{\pi }{4}} \right) = \cos \left( 0 \right) - c$
Here, the angle $\dfrac{\pi }{4}$ is written in degrees as $45^\circ$
Now, we know the value of cosine function for angles $45^\circ$ and $0^\circ$
Substituting, those values in the above equation,
$\Rightarrow \dfrac{1}{{\sqrt 2 }} = 1 - c$
Now, for rationalizing on the left-hand side of the equation, we multiply the numerator and the denominator by $\sqrt 2$ .
$\Rightarrow \dfrac{{1 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} = 1 - c$
$\Rightarrow \dfrac{{\sqrt 2 }}{2} = 1 - c$
Adding $c$ on both sides of the equation,
$\Rightarrow \dfrac{{\sqrt 2 }}{2} + c = 1 - c + c$
$\Rightarrow \dfrac{{\sqrt 2 }}{2} + c = 1$
Subtracting $\dfrac{{\sqrt 2 }}{2}$ from both sides,
$\Rightarrow \dfrac{{\sqrt 2 }}{2} + c - \dfrac{{\sqrt 2 }}{2} = 1 - \dfrac{{\sqrt 2 }}{2}$
$\Rightarrow c = 1 - \dfrac{{\sqrt 2 }}{2}$
Substituting the value of $c$ in the general solution
$\Rightarrow \cos y = \cos x - \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right)$
 $\Rightarrow \cos y = \cos x - 1 + \dfrac{{\sqrt 2 }}{2}$

$\cos y = \cos x - 1 + \dfrac{{\sqrt 2 }}{2}$ is the particular solution of the given derivatives.

Note: Here, the particular solution means the equation is satisfied by a particularly unique value of $x$ and $y$ . While the general solution can have more than one solution. For considering the values for a particular solution, we have to remember that the value of the independent variable is inside the bracket, while the value of the dependent variable is on the opposite side of the equation.