Given:
Given the differential equation
[tex]y^{\prime}^{\prime}=\sin x[/tex]
having the general solution y = -sin x + Ax + B.
Required: Particular solution satisfying the initial conditions
[tex]y(\frac{\pi}{2})=0\text{ and }y^{\prime}(\frac{\pi}{2})=-2[/tex]
Explanation:
Plug the first initial condition to the general solution.
[tex]\begin{gathered} 0=-\sin(\frac{\pi}{2})+A\cdot\frac{\pi}{2}+B \\ A\cdot\frac{\pi}{2}+B=1\text{ ... \lparen1\rparen} \end{gathered}[/tex]
Now, find the derivative of the general solution.
[tex]y^{\prime}(x)=-\cos x+A[/tex]
Plug the second initial condition.
[tex]\begin{gathered} -2=-\cos(\frac{\pi}{2})+A \\ A=-2 \end{gathered}[/tex]
Substitute the value of A in equation (1).
[tex]\begin{gathered} -2\cdot\frac{\pi}{2}+B=1 \\ B=1+\pi \end{gathered}[/tex]
Now, substitute the values of both A and B into the general solution of the differential equation.
[tex]y=-\sin x-2x+1+\pi[/tex]
Final Answer: The particular solution is
[tex]y=-\sin x-2x+1+\pi[/tex]
