Derivation of Calculation Formula of First Order Linear Differential Equation

The form of the first order linear differential equation is as follows:

$$y^{\prime }+p(x)y=q(x)$$

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For the left side of the formula, it looks like the following formula, but not the same

$$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$$

$$y^{\prime }\cdot 1+y\cdot p(x)$$

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Here corresponds to $v$ seems to be unable to get a suitable one, but with$[e^{f(x)}{]}^{\prime }=e^{f(x)}\cdot f^{\prime }(x)$can be found to be similar, that is, the following formula

$$y^{\prime }\cdot e^{\int p(x)dx}+y\cdot e^{\int p(x)dx}\cdot p(x)$$

Come up with the same coefficients:

$$e^{\int p\left(x\right)dx}\cdot [y^{\prime }\cdot 1+y\cdot p(x)]$$

The same coefficient can be added to the right side of the original equal sign to get the whole formula (to be simplified)

$$e^{\int p\left(x\right)dx}\cdot [y^{\prime }\cdot 1+y\cdot p(x)]=e^{\int p\left(x\right)dx}\cdot q(x)$$

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So simplify to get:

$$[y\cdot e^{\int p(x)dx}{]}^{\prime }=e^{\int p(x)dx}\cdot q(x)$$

The left side is derivation, and the product is returned, and the integral sign is added to the right side:

$$y\cdot e^{\int p(x)dx}=\int e^{\int p\left(x\right)dx}\cdot q(x)dx+C$$

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Convert the y coefficient to 1 to get the final conclusionInterpretation formula：

$$y=e^{-\int p\left(x\right)dx}\cdot \int e^{\int p\left(x\right)dx}\cdot q\left(x\right)dx+C$$

And the general solution is all solutions