Question

Solve the initial value problem

$\frac{dy}{dx}=y^2\sin \left(x\right),y\left(\mathrm{\pi }\right)=1$

Short answer

The solution is$y\left(x\right)=\frac{1}{2+\cos \left(x\right)}$

Step-by-step solution

Step 1. Given information 

Given derivation is$\frac{dy}{dx}=y^2\sin \left(x\right),y\left(\mathrm{\pi }\right)=1$

Step 2. Solve the derivation 

$$\begin{gathered} \int \frac{1}{y^2}dy=\int \sin \left(x\right)dx \\ -\frac{1}{y}=-\cos \left(x\right)+c \\ y=\frac{1}{\cos \left(x\right)-c} \\ y\left(\mathrm{\pi }\right)=1 \\ 1=\frac{1}{-1-c} \\ -1-c=1 \\ c=-2 \\ substitutethecvalueinequation, \\ thesolutionisy\left(x\right)=\frac{1}{2+\cos \left(x\right)} \end{gathered}$$

Step 3. The solution 

The solution is$y\left(x\right)=\frac{1}{2+\cos \left(x\right)}$