[Math] The Linear First-Order Differential Equation

Form

$$ y' + p(x)y = q(x) $$

Solving Steps

$$
\begin{split}

y' + p(x)y &= q(x) \\
\text{Integrating Factor: } & e^{ \int p(x) dx } \\
y'e^{ \int p(x) dx } + p(x)ye^{ \int p(x) dx } &= q(x) e^{ \int p(x) dx } \\
\dfrac{d}{dx} \left ( y e^{ \int p(x) dx } \right ) &= q(x) e^{ \int p(x) dx } \\
y e^{ \int p(x) dx } &= \int q(x) e^{ \int p(x) dx } dx + C \\
y &= \cfrac{ \int q(x) e^{ \int p(x) dx } dx + C }{e^{ \int p(x) dx }}\\

\end{split}
$$

Examples

Ex. 1

$$
\begin{split}

y' - 2xy &= x \\
\text{Integrating Factor: } & e^{ \int -2 x dx } = e^{ -x^4 } \\
e^{-x^2} \left ( y' - 2 x y \right ) &= x e^{-x^2} \\
e^{-x^2} \cdot y' - 2 x e^{-x^2} y &= x e^{-x^2} \\
\left ( e^{-x^2} y \right )' &= x e^{-x^2} \\
e^{-x^2} y &= \int x e^{-x^2} dx + C \\
e^{-x^2} y &= - \dfrac{1}{2} e^{-x^2} + C \\
y &= \cfrac{ - \dfrac{1}{2} e^{-x^2} + C }{e^{-x^2}} \\

\end{split}
$$

Reference

https://www.youtube.com/watch?v=Et4Y41ZNyao

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