Separation of variables [10.FT]
Using separation of variables
To use the method:
- Write $y'(x)=\frac{dy}{dx}$.
- Pretend that you can do algebra with things like $dy$ and $dx$.
- Re-arrange to get $dy$ and all the $y$-related things on one side of the equation and $dx$ and all the $x$-related things on the other side.
- Integrate and solve for $y$!
Solve a familiar diff eq:
$$\frac{dy}{dx}=y$$ Re-arrange this to: $$\frac 1y dy = dx$$
Integrate both sides: $$\begineq \int\frac 1y dy =& \int dx\\ \ln y =& x + B\\ \endeq $$
Now, solve for $y$: $$\begineq e^{\ln y} =& e^{x + B}\\ y =& e^x e^B \endeq $$ Since $B$ was a constant, $e^B$ is just another constant. Let's call it $C$, and now the solution is:
$$y=Ce^x$$
Justification for separation of variables
Actually, what's going on is substitution (chain rule) to do this. Let's say that: $$\frac{dy}{dx}=f(x)g(y)=\frac{f(x)}{h(y)}.$$ Where $h(y)=1/g(y)$ will be fine as long as $g(y)\neq 0$.
Then we can use regular old algebra to multiply both sides by $h(y)$: $$h(y)\frac{dy}{dx} = f(x)$$
Let's think of $y=y(x)$ as a function of $x$, then we can write this as $$\begineq h(y(x))\,y'(x) =& f(x)\\ \int h(y(x))\,y'(x)\,dx =& \int f(x)\,dx \endeq $$
We can solve the integral on the left side using substitution. Instead of "$w(x)$" we'll use $y(x)$ as our substitution, because we already see under the integral $y'(x)\,dx=dy$, so:
$$\int h(y)\,dy = \int f(x)\,dx.$$
Unfortunately, not all differential equations can be solved by the method of separation of variables!
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