# Ordinary Differential Equations/Substitution 2

Substitution methods are really applicable anywhere you can find a differential equation. However, there's very few instances where you will always give a certain substitution. You generally pick one and plug it in as needed. So I'll give situations where you could use a substitution method, although you may later learn better methods.

## Parametric equations

One time where you may need it is when solving parametric equations. Lets say we're given functions for velocity in two dimensions- $v_{x}(t)$ and $v_{y}(t)$. If we want to solve for $y(x)$, you have to divide ${\frac {v_{y}}{v_{x}}}$. This works out to be ${\frac {\frac {dv_{y}}{dx}}{\frac {dv_{x}}{dx}}}={\frac {dy}{dx}}$. When you do this, you will frequently (although not always) get a chance to use ${\frac {y}{x}}$ substitution.

### Constant velocity

Lets say we're swimming across a river with constant velocity $v_{0}$. The river has no current. We start swimming at an angle of $\theta$ with respect to the shore. Solve for $y(x)$
The first thing we need to do is break the velocity into x and y components. This is fairly simple.
    $v_{x}=v_{0}cos(\theta )$
    $v_{y}=v_{0}sin(\theta )$
Using simple trig, we can remove the theta.
    $v_{x}=v_{0}{\frac {x}{\sqrt {x^{2}+y^{2}}}}$
    $v_{y}=v_{0}{\frac {y}{\sqrt {x^{2}+y^{2}}}}$
Now we divide the two to find ${\frac {dy}{dx}}$.
    ${\frac {dy}{dx}}={\frac {v_{0}{\frac {y}{\sqrt {x^{2}+y^{2}}}}}{v_{0}{\frac {x}{\sqrt {x^{2}+y^{2}}}}}}={\frac {y}{x}}$
Now this is simple to solve separably. It could also be solved via substitution. This is a trivial example, but it can be made more complicated.

### Motion against a current

Imagine the same swimmer. Now there is a current with speed r going straight up the river (positive y direction). How does this change our example?
The x component is still the same.
    ${\frac {dx}{dt}}=v_{0}cos(\theta )={\frac {v_{0}x}{\sqrt {x^{2}+y^{2}}}}$
And in the y direction we also have a term due to the current.
    ${\frac {dy}{dt}}=v_{0}sin(\theta )+r={\frac {v_{0}y}{\sqrt {x^{2}+y^{2}}}}+r$
You can get ${\frac {dy}{dx}}$ by dividing the two equations
    ${\frac {dy}{dx}}={\frac {y}{x}}+{\frac {r{\sqrt {x^{2}+y^{2}}}}{v_{0}x}}$
We can move the x into the root to simplify the equation a bit
    ${\frac {dy}{dx}}={\frac {y}{x}}+{\frac {r{\sqrt {{\frac {x^{2}}{x^{2}}}+{\frac {y^{2}}{x^{2}}}}}}{v_{0}}}$
    ${\frac {dy}{dx}}={\frac {y}{x}}+{\frac {r}{v_{0}}}{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}$
Well, this complicated equation looks like a case for ${\frac {y}{x}}$ substitution.
    $v={\frac {y}{x}}$
    $y=vx$
    $y'=v+v'x$
    $v+v'x=v+{\frac {r}{v_{0}}}{\sqrt {1+v^{2}}}$
    $v'={\frac {r}{xv_{0}}}{\sqrt {1+v^{2}}}$
That looks like a nice, easily solved separable equation. Let solve it.
    ${\frac {dv}{\sqrt {1+v^{2}}}}={\frac {r}{xv_{0}}}$
    $\int {\frac {dv}{\sqrt {1+v^{2}}}}=\int {\frac {r}{xv_{0}}}$
The left end is an ugly integral. Just trust me on it.
    $ln(v+{\sqrt {1+v^{2}}})={\frac {r}{v_{0}}}ln(x)+C$
    $v+{\sqrt {1+v^{2}}}=e^{ln(x^{\frac {r}{v_{0}}})+C}$
    $v+{\sqrt {1+v^{2}}}=Cx^{\frac {r}{v_{0}}}$
Lets try to get rid of that root. Isolate it, and square both sides.
    ${\sqrt {1+v^{2}}}=Cx^{\frac {r}{v_{0}}}-v$
    $1+v^{2}=Cx^{2{\frac {r}{v_{0}}}}-2vCx^{\frac {r}{v_{0}}}+v^{2}$
    $2vCx^{\frac {r}{v_{0}}}=C^{2}x^{2{\frac {r}{v_{0}}}}-1$
    $v={\frac {Cx^{\frac {r}{v_{0}}}}{2}}-{\frac {1}{2Cx^{\frac {r}{v_{0}}}}}$
Plugging in for v, we get
    ${\frac {y}{x}}={\frac {Cx^{\frac {r}{v_{0}}}}{2}}-{\frac {1}{2Cx^{\frac {r}{v_{0}}}}}$
We can solve for y by multiplying through by x
    $y={\frac {Cx^{{\frac {r}{v_{0}}}+1}}{2}}-{\frac {x}{2Cx^{\frac {r}{v_{0}}}}}$
    $y={\frac {C}{2}}x^{{\frac {r}{v_{0}}}+1}-{\frac {1}{2C}}x^{{\frac {-r}{v_{0}}}+1}$
This complicated equation does make sense- the bigger the current, the further you go in the y direction as a portion of the x.
If you ever find an equation this evil in real life, do yourself a favor and buy a computer program to solve it.