# Why this equation is correct?

I am going through the University of Pennsylvania's numerical analysis lectures, which can be found here. I am a bit confused with equation 1.1.9 at page 6-7.

The equation is $y'(x) + a(x)y(x) = b(x)$. The notes say that this can be rewritten as

$$a^{-A(x)}d(a^{A(x)}y(x)) = b(x)$$ where $A(x)$ is an antiderivative of $a(x)$. So, we have that $$y'(x) + a(x)y(x) = a^{-A(x)}d(a^{A(x)}y(x)).$$

Why is that? I can not go from the first equation to the other.

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This is deceptively anticlimactic, and in one phrase the answer is the product rule of differentiation, by which I mean that $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$.

Claim:

$$e^{-A(x)}\frac{d}{dx} \left(e^{A(x)}y(x))\right) = y'(x) + a(x)y(x)$$

Proof: let's take the derivative on the left.

$$\frac{d}{dx} \left( d^{A(x)}y(x)\right) = e^{A(x)}a(x)y(x) + e^{A(x)}y'(x),$$

so that multiplying by $e^{-A(x)}$ gives exactly what we want. $\diamondsuit$

In terms of differential equations, this sort of method is usually called an integrating factor.

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thanks a lot mixedmath. Very good answer :) – K. Stasko Jun 16 '14 at 20:50

You have not transcribed the notes correctly. The notes say to consider $$y'(x) + a(x) y(x) = b(x)$$

Then we rewrite this equation in the equivalent form

$$e^{-A(x)} \dfrac{d}{dx} \left( e^{A(x)} y(x) \right) = b(x)$$

Why is this correct? Because if we apply the derivative operator on the left hand side, we get

$$e^{-A(x)} \left(A'(x) e^{A(x)} y(x) + e^{A(x)} y'(x) \right) = A'(x)y(x) + y'(x) = b(x)$$

and recall $A'(x) = a(x)$.

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Thanks a lot A.E. – K. Stasko Jun 16 '14 at 20:51

This is called an integrating factor. If we multiply the whole equation by $\exp\left(\int a(x)\,dx\right)$, we get

$$e^{\int a(x)\,dx}y'(x)+a(x)e^{\int a(x)\,dx}y(x) = b(x)e^{\int a(x)\,dx}.$$

Let's consider $\left(e^{\int a(x)\,dx}y(x)\right)'$. By product rule, we get

$$\left(e^{\int a(x)\,dx}\right)'y(x) + e^{\int a(x)\,dx}y'(x).$$

Moreover,

$$\left(e^{\int a(x)\,dx}\right)' = e^{\int a(x)\,dx}\left(\int a(x)\,dx\right)' = a(x)e^{\int a(x)\,dx}.$$

Thus we can realize $\left(e^{\int a(x)\,dx}y(x)\right)'$ as $a(x)e^{\int a(x)\,dx}y(x)+e^{\int a(x)\,dx}y'(x).$ So then our overall expression is

$$\left(e^{\int a(x)\,dx}y(x)\right)' = e^{\int a(x)\,dx}b(x).$$

Multiplying both sides by $e^{-\int a(x)\,dx}$ gives the result. See my answer here for a much more detailed exposition on this matter.

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Thanks a lot Cameron – K. Stasko Jun 16 '14 at 20:51