Differential Equations/Substitution 2

From Wikibooks, the open-content textbooks collection

Jump to: navigation, search

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.

[edit] 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- vx(t) and vy(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.

[edit] Constant velocity

Lets say we're swimming across a river with constant velocity v0. The river has no current. We start swimming at an angle of θ 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.

vx = v0cos(θ)
vy = v0sin(θ)

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.


[edit] 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_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.

I won't torture you by doing more with this, its too ugly. If you ever find an equation this evil in real life, do yourself a favor and buy a computer program to solve it.

Retrieved from "http://en.wikibooks.org/wiki/Differential_Equations/Substitution_2"