# f(x)=e^(ux) where u is a root of a(x^2)+bx+c=0. What is af ''(x) +bf '(x) +cf(x) in fully simplified form.

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### 1 Answer

You need to differentiate the function with respect to x, two times, `such that: f'(x) = (e^(ux))'`

`f'(x) = e^(ux)*(ux)'`

`f'(x) = u*e^(ux)`

`f''(x) = u^2*e^(ux)`

You need to substitute `u*e^(ux)` for f'(x) and `u^2*e^(ux)` for f''(x) in `af ''(x) +bf '(x) +cf(x)` such that:

`a*u^2*e^(ux) + b*u*e^(ux)+ ce^(ux)`

You need to factor out `e^(ux)` such that:

`e^(ux)*(a*u^2 + b*u + c)`

The problem provides the information `that u is a root for a*x^2 + b*x + c = 0` , hence, substituting u for x in equation, the equation is equal to zero such that:

`e^(ux)*(a*u^2 + b*u + c) = e^(ux)*(0) = 0`

**Hence, evaluating `af ''(x) +bf '(x) +cf(x)` under given conditions yields `af ''(x) +bf '(x) +cf(x) = 0.` **