Question

Solve the initial value problems: 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

Answer 1

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 }$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$