Determine if the given function y = f(x) is a solution of the accompanying differential equation.
-Differential equation: $2 xy ^ { \prime } + 2 y = \cos x$ Initial condition: $\mathrm { y } ( \pi ) = 0$
Solution candidate: $y = \frac { \sin x } { 2 x }$

Question

Quizplus · Accepted Answer

First, we need to take the derivative of the solution candidate, which is:

$$y' = \frac{2x\cos x - \sin x}{4x^2}$$

Now, we can substitute both the solution candidate and its derivative into the differential equation:

\begin{align*}
2xy' + 2y &= \cos x \
2x\left(\frac{2x\cos x - \sin x}{4x^2}ight) + 2\left(\frac{\sin x}{2x}ight) &= \cos x \
\cos x - \frac{\sin x}{x} + \frac{\sin x}{x} &= \cos x \
\end{align*}

The left-hand side simplifies to just $\cos x$, which is indeed equal to the right-hand side. Therefore, the given solution candidate is a solution to the differential equation.

Now, we need to check if the initial condition is satisfied:

$$y(\pi) = \frac{\sin(\pi)}{2\pi} = 0$$

Since $\sin(\pi) = 0$, we see that the initial condition is satisfied as well. Therefore, the given function is a solution to the differential equation with the given initial condition.

Determine If the Given Function Y = F(x) Is a Solution