the equation $y = 2e^{6x}-5$ is a particular solution to which of the following differential equations?\n(a)…

Explanation:

Step1: Find the first - derivative of $y$

Given $y = 2e^{6x}-5$, using the derivative rule $\frac{d}{dx}(e^{ax})=ae^{ax}$, we have $y'=\frac{d}{dx}(2e^{6x}-5)=2\times6e^{6x}=12e^{6x}$.

Step2: Check option A

Substitute $y$ and $y'$ into the left - hand side of the equation $y' - 6y-30$. $y' - 6y-30=12e^{6x}-6(2e^{6x}-5)-30$. Expand the expression: $12e^{6x}-12e^{6x}+30 - 30=0$.

Step3: Check option B

Substitute $y$ and $y'$ into the left - hand side of the equation $2y' - 12y + 5$. $2y' - 12y+5=2\times12e^{6x}-12(2e^{6x}-5)+5$. Expand: $24e^{6x}-24e^{6x}+60 + 5=65\neq0$.

Step4: Check option C

Find the second - derivative $y''$. Since $y' = 12e^{6x}$, then $y''=\frac{d}{dx}(12e^{6x})=72e^{6x}$. Substitute $y$, $y'$ and $y''$ into the left - hand side of the equation $y''-5y'-6y$. $y''-5y'-6y=72e^{6x}-5\times12e^{6x}-6(2e^{6x}-5)$. Expand: $72e^{6x}-60e^{6x}-12e^{6x}+30=30\neq0$.

Step5: Check option D

Substitute $y$, $y'$ and $y''$ into the left - hand side of the equation $y''-2y'+y + 5$. $y''-2y'+y + 5=72e^{6x}-2\times12e^{6x}+2e^{6x}-5 + 5$. Combine like terms: $(72 - 24+2)e^{6x}=50e^{6x}\neq0$.

Answer:

A. $y' - 6y-30 = 0$