Prove that y = (e^5x+2)/(e^x) satisfies y'''-13y'-12y = 0

Prove that $y = \dfrac{e^{5x}+2}{e^{x}}$ satisfies $y'''-13y'-12y = 0$.

Step 1: Split the quotient

Dividing each term by $e^{x}$ removes the fraction entirely: $y = \frac{e^{5x}}{e^{x}}+\frac{2}{e^{x}} = e^{4x}+2e^{-x}$.

Step 2: Differentiate three times

\[ y' = 4e^{4x}-2e^{-x}, \qquad y'' = 16e^{4x}+2e^{-x}, \qquad y''' = 64e^{4x}-2e^{-x}. \]

Step 3: Substitute and collect

The two exponentials are independent functions, so the coefficient of each must be checked separately: $e^{4x}:\quad 64-13\cdot4-12\cdot1 = 64-52-12 = 0$.

\[ e^{-x}:\quad -2-13\bigl(-2\bigr)-12\cdot2 = -2+26-24 = 0. \]

Both vanish, so $y'''-13y'-12y = 0$ identically.

Step 4: Why those two numbers work

The two arithmetic checks are exactly the statement that $r = 4$ and $r = -1$ satisfy the characteristic equation

\[ r^{3}-13r-12 = 0. \]

The cubic factors completely: $r^{3}-13r-12 = \bigl(r-4\bigr)\bigl(r+1\bigr)\bigl(r+3\bigr)$.

so the roots are $4,\,-1,\,-3$ and the general solution is $C_{1}e^{4x}+C_{2}e^{-x}+C_{3}e^{-3x}$. The given function is the particular member with $C_{1} = 1,\ C_{2} = 2,\ C_{3} = 0$.

Step 5: Verification

\[r = 4:\ 64-52-12 = 0 \;\;\checkmark; \qquad r = -1:\ -1+13-12 = 0 \;\;\checkmark; \qquad r = -3:\ -27+39-12 = 0 \;\;\checkmark\]

\[\text{Sum of roots } 4-1-3 = 0, \text{ matching the absent } r^{2} \text{ term} \;\;\checkmark\]

\[x = 0:\ y = 3,\ y' = 2,\ y''' = 62; \quad 62-13\cdot2-12\cdot3 = 62-26-36 = 0 \;\;\checkmark\]