Determine if the given function y = f(x) is a solution of the accompanying differential equation.
-Differential equation: $6 x y ^ { \prime } + 6 y = \cos x$
Initial condition: $y ( \pi ) = 0$
Solution candidate: $y = \frac { 6 \sin x } { x }$

Question

Quizplus · Accepted Answer

To verify if $y = \frac{6\sin x}{x}$ is a solution, we need to check if it satisfies the differential equation $6xy' + 6y = \cos x$. Differentiating $y$ with respect to $x$ gives $y' = \frac{6x\cos x - 6\sin x}{x^2}$. Substituting $y$ and $y'$ into the differential equation yields $6x\left(\frac{6x\cos x - 6\sin x}{x^2}\right) + 6\left(\frac{6\sin x}{x}\right) = \cos x$, which simplifies to $\cos x = \cos x$, indicating the solution does satisfy the differential equation. However, the initial condition $y(\pi) = 0$ is not satisfied by $y = \frac{6\sin x}{x}$ because $y(\pi) = \frac{6\sin \pi}{\pi} = 0$, which actually does satisfy the initial condition. My initial statement was incorrect; the function does satisfy both the differential equation and the initial condition.

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