Find the particular solution to y "= sin(x) given the general solution y = -sin(x) + Ax + B and the initial conditions

Question

asked Jan 12, 2023 98.3k views

1 Answer

Josha Inglis · Answer 1 · 2023-01-19T10:34:48+0000

Given:

Given the differential equation

$y^(\prime)^(\prime)=\sin x$

having the general solution y = -sin x + Ax + B.

Required: Particular solution satisfying the initial conditions

$y((\pi)/(2))=0\text{ and }y^(\prime)((\pi)/(2))=-2$

Step-by-step explanation:

Plug the first initial condition to the general solution.

$\begin{gathered} 0=-\sin((\pi)/(2))+A\cdot(\pi)/(2)+B \\ A\cdot(\pi)/(2)+B=1\text{ ... \lparen1\rparen} \end{gathered}$

Now, find the derivative of the general solution.

$y^(\prime)(x)=-\cos x+A$

Plug the second initial condition.

$\begin{gathered} -2=-\cos((\pi)/(2))+A \\ A=-2 \end{gathered}$

Substitute the value of A in equation (1).

$\begin{gathered} -2\cdot(\pi)/(2)+B=1 \\ B=1+\pi \end{gathered}$

Now, substitute the values of both A and B into the general solution of the differential equation.

$y=-\sin x-2x+1+\pi$

Final Answer: The particular solution is

$y=-\sin x-2x+1+\pi$