Differential Equations/Substitution 1

First-Order Differential Equations

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.

General substitution procedure

All substitution methods use the same general procedure:

1. Take a term of the equation and replace it with a variable v. The new variable must replace all instances of the variable y.
2. 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.
   
3. Plug in $\frac{dv}{dx}$ and solve for $v$.
4. Plug $v$ into the original term replaced, and solve for $y$.

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).

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

I highly suggest you do not memorize these equation, and instead remember the method of solving the problem. The final equation is rather obscure and easy to forget, but if you know the method you can always solve it. It will also help you use other substitution methods.

Example 1

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

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

$v=y+x+3$

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).

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

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

"Homogeneous" substitution

If we're given an equation in the form $\frac{dy}{dx}=F\left(\frac{y}{x}\right)$, we can replace $\frac{y}{x}$ with v to make it easier to solve. Confusingly, this kind of DE can be called homogeneous, but it does not mean the same as the definition of homogeneous in Form of a Differential Equation.

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

$v(x,y)=\frac{y}{x}$

$y=vx$

Now we need to find v':

$\frac{dy}{dx}=v+\frac{dv}{dx}x$

Plug back into the original equation

$v+x\frac{dv}{dx}=F(v)$

$\frac{dv}{dx}=\frac{F(v)-v}{x}$

Solve for v(x), then plug into the equation of v to get y

$y(x)=xv(x)$

Again, don't memorize the equation. Remember the general method, and apply it.

Example 2

$\frac{dy}{dx}=5\frac{y}{x}+3\frac{x}{y}$

Let's use $v=\frac{y}{x}$. Solve for y'(x,v,v')

$y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now plug into the original equation

$v+x\frac{dv}{dx}=5v+\frac{3}{v}$

$x\frac{dv}{dx}=4v+\frac{3}{v}$

$v\frac{dv}{dx}=\frac{(4v^2+3)}{x}$

Solve for v

$\frac{vdv}{4v^2+3}=\frac{dx}{x}$

$\int \frac{vdv}{4v^2+3}=\int \frac{dx}{x}$

$\frac{1}{8}\ln (4v^2+3)=\ln (x)$

$4v^2+3=e^{8\ln (x)}$

$4v^2+3=e^{\ln (x^8)}$

$4v^2+3=x^8$

$v^2=\frac{x^8-3}{4}$

Plug into the definition of v to get y.

$y=vx$

$y^2=v^2{x}^2$

$y^2=\frac{x^{10}-3x^2}{4}$

We leave it in $y^2$ form, since solving for y would lose information.

Example 3

$\frac{dy}{dx}=\frac{x}{\sin (\frac{y}{x})}+\frac{y}{x}$

Lets use $v=\frac{y}{x}$ again. Solve for $y^{\prime }(x,v,v^{\prime })$

$y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now plug into the original equation

$v+x\frac{dv}{dx}=\frac{x}{\sin (v)}+v$

$x\frac{dv}{dx}=\frac{x}{\sin (v)}$

$\sin (v)dv=dx$

Solve for v:

$\int \sin (v)dv=\int dx$

$-\cos v=x+C$

$v=\arccos (-x+C)$

Use the definition of v to solve for y.

$y=vx$

$y=\arccos (-x+C)x$

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.

Example 4

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

This equation isn't seperable, 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':

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

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

Plug into the original equation

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

Solve for v

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

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

$\ln (v+1)=x+C$

$v=C{e}^x-1$

Now plug in and get y

$C{e}^x-1=y^2+x-1$

$y^2=C{e}^x-x$

Pretty easy after using that substitution. Keep this method in mind, you will use this for more complex equations.