Differential Equations/Substitution 1

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As we saw in a previous example, sometimes even though an equation isn't separable in its original form, it can be factored into a form where it is. Another way you can turn non-separable equations into separable ones is to use substitution methods.

[edit] General substitution procedure

All substitution methods use the same general procedure:

Take a term of the equation and replace it with a variable v. The new variable must replace all instances of the variable y.
Solve for $\frac{dy}{dx}$ in terms of $v$ and $\frac{dv}{dx}$ . To do this, take the equation $v = f (x, y)$ where $f$ is the term you replaced and take its derivative.
Plug in $\frac{dv}{dx}$ and solve for $v$ .
Plug $v$ into the original term replaced, and solve for $y$ .

[edit] Constant coefficient substitution

Lets say we have an equation with a term f(x)=ay+bx+c, such as

$\frac{dy}{dx}=G(ay+bx+c).$

This is non-separable. But we can sometimes solve these equations by replacing the term with v.

First, we define v(x,y) and find v'(x,y,y').

$v(x,y)=ay+bx+c \,$

$\frac{dv}{dx}=a\frac{dy}{dx}+b$

Next, we solve for y'(x,v,v'):

$\frac{dy}{dx}=\frac{\frac{dv}{dx}-b}{a}$

Now plug into the original equation, and get it into the form $\frac{dv}{dx}=f(v)$

$\frac{dy}{dx}=G(ay+bx+c)$

$\frac{\frac{dv}{dx}-b}{a}=G(v)$

$\frac{dv}{dx}=aG(v)+b$

Solve for v:

$\frac{dv}{dx}=aG(v)+b$

$\frac{dv}{aG(v)+b}= dx$

$\int \frac{dv}{aG(v)+b}= \int dx$

$\int \frac{dv}{aG(v)+b}= x+D$

Once you have v(x), plug back into the definition of v(x) to get y(x).

$y(x)=\frac{v(x)-c-bx}{a}$

It is highly suggested that one should not memorize this equation, and instead remember the method of solving the problem. The final equation is rather obscure and easy to forget, but if one knows the method, he/she can always solve it. It will also help if one uses other substitution methods.

[edit] Example 1

$\frac{dy}{dx}=(x+y+3)^2$

Lets replace the quantity being raised to a power with v.

$v=x+y+3 \,$

Now lets find v'.

$\frac{dv}{dx}=\frac{dy}{dx}+1$

Solve for y'

$\frac{dy}{dx}=\frac{dv}{dx}-1$

Plug in for y and y':

$\frac{dv}{dx}-1=v^2$

$\frac{dv}{dx}=v^2+1$

Now we solve for v, using the methods we learned in Separable Variables:

$\frac{dv}{v^2+1}=dx$

$\int \frac{dv}{v^2+1}=\int dx$

$\tan^{-1}(v)=x+C \,$

$v=\tan(x+C) \,$

Now that we have v(x), plug back in and find y(x).

$y+x+3=\tan(x+C) \,$

$y=\tan(x+C)-x-3 \,$

[edit] Other methods

These are not the only possible substitution methods, just some of the more common ones. Substitution methods are a general way to simplify complex differential equations. If you ever come up with a differential equation you can't solve, you can sometimes crack it by finding a substitution and plugging in. Just look for something that simplifies the equation. Remember that between v and v' you must eliminate the y in the equation.

[edit] Example 4

$2y\frac{dy}{dx}=y^2+x-1$

This equation isn't separable, and none of the methods we previously used will quite work. Let's use a custom substitution of v=y²+x-1. Solve for v':

$\frac{dv}{dx}=2y\frac{dy}{dx}+1$