$(4$ pts$)16)$ Briefly explain what is wrong with the following attempt to solve the linear differential equation $y^{\text{'}}+3y=sinx$

For this linear DE$,P(x)=3$ and $Q(x)=sinx$

The integrating factor is $I(x)=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}3dx}=e^{3x}$

Multiplying both sides by $e^{3x}$ then gives $e^{3x}{y}^{\text{'}}+3e^{3x}y=sinx$

The left side is a product rule with $F=e^{3x}$ and $S=y$

so integrating both sides gives $e^{3x}y=-cosx+C$

Finally, dividing by $e^{3x}$ gives $y=-\frac{cosx}{e^{3x}}+C{e}^{-3x}$