Last modified on 27 November 2010, at 05:43

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 equationsEdit

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 velocityEdit

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_0cos(\theta)
v_y=v_0sin(\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 substituion. This is a trivial example, but it can be made more complicated.


Motion against a currentEdit

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_0cos(\theta)=\frac{v_0x}{\sqrt{x^2+y^2}}

And in the y direction we also have a term due to the current.

\frac{dy}{dt}=v_0sin(\theta)+r=\frac{v_0y}{\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_0x}

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^2x^{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.