Suppose f(x) = e^{ax} + e^{bx}, where a neq b | vedclass

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(C) Given $f(x) = e^{ax} + e^{bx}.$Then $f'(x) = ae^{ax} + be^{bx}$ and $f''(x) = a^2e^{ax} + b^2e^{bx}.$Substituting these into the given differentia

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$-15$

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(C) Given $f(x) = e^{ax} + e^{bx}.$Then $f'(x) = ae^{ax} + be^{bx}$ and $f''(x) = a^2e^{ax} + b^2e^{bx}.$Substituting these into the given differential equation $f''(x) - 2f'(x) - 15f(x) = 0,$$(a^2e^{ax} + b^2e^{bx}) - 2(ae^{ax} + be^{bx}) - 15(e^{ax} + e^{bx}) = 0.$Grouping the terms with $e^{ax}$ and $e^{bx},$$(a^2 - 2a - 15)e^{ax} + (b^2 - 2b - 15)e^{bx} = 0.$Since $e^{ax}$ and $e^{bx}$ are linearly independent functions,their coefficients must be zero:$a^2 - 2a - 15 = 0$ and $b^2 - 2b - 15 = 0.$Solving the quadratic equation $t^2 - 2t - 15 = 0,$$(t - 5)(t + 3) = 0,$which gives $t = 5$ or $t = -3.$Since $a 
eq b,$ we must have $a = 5$ and $b = -3$ (or vice versa).Therefore,the product $ab = 5 	imes (-3) = -15.$

Suppose $f(x) = e^{ax} + e^{bx},$ where $a \neq b,$ and that $f''(x) - 2f'(x) - 15f(x) = 0$ for all $x.$ Then the product $ab$ is equal to

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