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Show that equations of the form $\ (b_1 (x) y'+ b_0 (x) y)' = f(x) $ are exact.

I do not understand how to prove this when the whole LHS is the derivative and there is a derivative inside a derivative.

we are given: $ a_2 (x) y''+ a_1 (x) y' + a_0 (x) y = f(x) $ is an exact equation if $ a_0 - a_1 ' + a_2 '' = 0 $ how do we use this hint to show whether $\ (b_1 (x) y'+ b_0 (x) y)' = f(x) $ is an exact equation.

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Well, you can always expand the derivative and get

$$ b_1'(x)y' + b_1(x)y'' + b_0'(x)y + b_0(x)y' = b_1(x)y'' + (b_1'(x) + b_0(x))y' + b_0'(x)y = f(x). $$

Then, if you set $a_2(x) = b_1(x), a_1(x) = b_1'(x) + b_0(x)$ and $a_0(x) = b_0'(x)$ you have

$$ a_0(x) - a_1'(x) + a_2''(x) = b_0'(x) - b_1''(x) - b_0'(x) + b_1''(x) = 0.$$

The reason that exact equations are interesting in the first place is that they can be written as

$$ (b_1(x)y' + b_0(x)y)' = f(x). $$

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Apply the outer derivative, and use the product rule. \begin{align} (b_1 (x) y'+ b_0 (x) y)' = f(x) \\ \\ b_1' (x) y'+b_1' (x) y''+ b_0 '(x) y + b_0 '(x) y' = f(x) \end{align}

Can you take it from here?

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  • $\begingroup$ i wasnt aware you can use product rule in such way, so it makes sense now! $\endgroup$ – ISuckAtMathPleaseHELPME Dec 1 '16 at 2:43

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