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Solving Separable Differential Equations

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How do you solve separable differential equations with initial conditions?

A separable equation typically looks like:

$\frac{d y}{d x} = \frac{g (x)}{f (y)}$ ${dy}/{dx}={g(x)}/{f(y)}$ .

by multiplying by $d x$ $dx$ and by $f (y)$ $f(y)$ to separate $x$ $x$ 's and $y$ $y$ 's,

$\Rightarrow f (y) d y = g (x) d x$ $Rightarrow f(y)dy=g(x)dx$

by integrating both sides,

$\Rightarrow \int f (y) d y = \int g (x) d x$ $Rightarrow int f(y)dy=int g(x)dx$ ,

which gives us the solution expressed implicitly:

$\Rightarrow F (y) = G (x) + C$ $Rightarrow F(y)=G(x)+C$ ,

where $F$ $F$ and $G$ $G$ are antiderivatives of $f$ $f$ and $g$ $g$ , respectively.

For an example of a separable equation with an initial condition, please watch this video:

Wataru · · Oct 1 2014
How do you solve separable differential equations?

A separable equation typically looks like:
$\frac{d y}{d x} = \frac{g (x)}{f (y)}$ ${dy}/{dx}={g(x)}/{f(y)}$ .

By multiplying by $d x$ $dx$ and by $g (y)$ $g(y)$ to separate $x$ $x$ 's and $y$ $y$ 's,
$\Rightarrow f (y) d y = g (x) d x$ $Rightarrow f(y)dy=g(x)dx$

By integrating both sides,
$\Rightarrow \int f (y) d y = \int g (x) d x$ $Rightarrow int f(y)dy=int g(x)dx$ ,
which gives us the solution expressed implicitly:

$\Rightarrow F (y) = G (x) + C$ $Rightarrow F(y)=G(x)+C$ ,
where $F$ $F$ and $G$ $G$ are antiderivatives of $f$ $f$ and $g$ $g$ , respectively.

For more details, please watch this video:

Wataru · 1 · Sep 19 2014

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Applications of Definite Integrals

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