Respuesta :
Answer:
[tex]-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}[/tex]
Step-by-step explanation:
Given the initial value problem [tex]\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\[/tex] subject to y(π) = 1. To solve this we will use the variable separable method.
Step 1: Separate the variables;
[tex]\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}[/tex]
[tex]-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}[/tex]
The solution to the initial value problem will be;
[tex]-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}[/tex]
Using separation of variables, it is found that the solution of the initial value problem is:
- [tex]\frac{y^2}{2} + \ln{y} + \pi - \frac{1}{2} + \theta\cos{\theta} - \sin{\theta} = 0[/tex]
The differential equation is given by:
[tex]\frac{1}{\theta}\left(\frac{dy}{d\theta}\right) = \frac{y\sin{\theta}}{y^2 + 1}[/tex]
Separation of variables:
Applying separation of variables, we have that:
[tex]\frac{y^2 + 1}{y}dy = \theta\sin{\theta}d\theta[/tex]
[tex]\int \frac{y^2 + 1}{y}dy = \int \theta\sin{\theta}d\theta[/tex]
The first integral is solved applying the properties, as follows:
[tex]\int \frac{y^2 + 1}{y}dy = \int y dy + \int \frac{1}{y} dy = \frac{y^2}{2} + \ln{y} + K[/tex]
- In which K is the constant of integration.
The second integral is solved using integration by parts, as follows:
[tex]u = \theta, du = d\theta[/tex]
[tex]v = \int \sin{\theta}d\theta = -\cos{\theta}[/tex]
Then:
[tex]\int \theta\sin{\theta}d\theta = uv - \int v du[/tex]
[tex]\int \theta\sin{\theta}d\theta = -\theta\cos{\theta} + \int \cos{\theta}d\theta[/tex]
[tex]\int \theta\sin{\theta}d\theta = -\theta\cos{\theta} + \sin{\theta}[/tex]
Then:
[tex]\frac{y^2}{2} + \ln{y} + K = -\theta\cos{\theta} + \sin{\theta}[/tex]
[tex]y(\pi) = 1[/tex] means that when [tex]\theta = \pi, y = 1[/tex], which is used to find K.
[tex]\frac{1}{2} + \ln{1} + K = -\pi\cos{\pi} + \sin{\pi}[/tex]
[tex]\frac{1}{2} + K = \pi[/tex]
[tex]K = \pi - \frac{1}{2}[/tex]
Then, the solution is:
[tex]\frac{y^2}{2} + \ln{y} + \pi - \frac{1}{2} = -\theta\cos{\theta} + \sin{\theta}[/tex]
[tex]\frac{y^2}{2} + \ln{y} + \pi - \frac{1}{2} + \theta\cos{\theta} - \sin{\theta} = 0[/tex]
To learn more about separation of variables, you can take a look at https://brainly.com/question/14318343
