[Math] Separable 1st Order Differential Equation

Form

1st ODE

$$ \dfrac{dy}{dx} = F \left ( x, y \right ) $$

Separable

$$
\begin{equation}
\begin{split}

\dfrac{dy}{dx} & = F \left ( x \right ) G \left ( y \right ) \\
\\
\dfrac{1}{G(y)} dy & = F(x) dx

\end{split}
\end{equation}
$$

Examples

Ex. 1

$$
\begin{equation}
\begin{split}

\left ( x^2 + 1 \right ) \dfrac{dy}{dx} & = xy \\
\dfrac{ \left ( x^2 + 1 \right ) }{ \left ( x^2 + 1 \right ) y } dy & =
\dfrac{xy}{ \left ( x^2 + 1 \right ) y } dx \\
\dfrac{1}{y} dy & = \left ( \dfrac{x}{x^2 + 1} \right ) dx \\
\int \dfrac{1}{y} dy & = \int \left ( \dfrac{x}{x^2 + 1} \right ) dx \\
\ln \mid y \mid + C_1 & = \frac{1}{2} \int \dfrac {1}{u} du \\
\ln \mid y \mid & = \frac{1}{2} \ln \mid x^2 + 1 \mid + C_2 \\
e^{ \ln \mid y \mid + C_1 } & = e^{ \frac{1}{2} \ln \mid x^2 + 1 \mid + C_2 } \\
\mid y \mid & = \left ( x^2 + 1 \right )^{ \frac{1}{2} } \cdot e^{ C_2 - C_1 } \\
\mid y \mid & = \left ( x^2 + 1 \right )^{ \frac{1}{2} } \cdot C

\end{split}
\end{equation}
$$

$$
\begin{equation}
\begin{split}

\text{let } u & = x^2 + 1 \\
dv & = 2x dx \\
\frac{1}{2} dv & = x dx \\
\\
\int \left ( \dfrac{x}{x^2 + 1} \right ) dx & =
\frac{1}{2} \int \frac{1}{u} du

\end{split}
\end{equation}
$$

General solution:
$$ \mid y \mid = \left ( x^2 + 1 \right )^{ \frac{1}{2} } \cdot C $$

Ex. 2

Initial Value Problem
$$
\begin{equation}
\begin{split}

\dfrac{dy}{dx} & = \dfrac{ -x }{ y e^{ x^2} } \\
y \cdot dy & = \dfrac{ -x }{ e^{ x^2} } \cdot dx \\
y \cdot dy & = -x e^{ - x^2 } \cdot dx \\
\int y \cdot dy & = \int -x e^{ - x^2 } \cdot dx \\
\dfrac{ y^2 }{ 2 } + C_1 & = \frac{1}{2} e^{ -x^2 } + C_2 \\
\dfrac{ y^2 }{ 2 } & = \frac{1}{2} e^{ -x^2 } + C \\
y^2 & = e^{ -x^2 } + C \\

\end{split}
\end{equation}
$$

General Solution:
$$ y^2 = e^{ -x^2 } + C $$

Particular Solution:
$$
\begin{equation}
\begin{split}

y^2 & = e^{ -x^2 } + C \text{ for } y(0) = 1 \\
1 & = 1 + C \\
C & = 0 \\
\\
y^2 & = e^{ -x^2 } + 0 \\
y & = \pm \sqrt{ e^{ -x^2 } } \\
\\
\because y &= 1 \text{ is positive } \\
\therefore y & = \sqrt{ e^{ -x^2 } } \\
y & = e^{ - \frac{ x^2 }{2} } \\

\end{split}
\end{equation}
$$

Reference

https://www.youtube.com/watch?v=nNHlSB6b1HU
https://www.youtube.com/watch?v=DL-ozRGDlkY

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