Math Help Forum - View Single Post

Chris L T521 · #1 May 13th, 2008, 01:12 AM

Once in a while, differential equation questions pop up, so I'm going to point out the various techniques on how to solve them:

1. Direct Integration:

If you have a differential equation in the form $\frac{dy}{dx}=f(x)$ , we can use direct integration to solve the DE.

Example 1:

Solve $\frac{dy}{dx}=3x^2$ .

To solve, simply integrate both sides of the equation:

$\int\frac{dy}{dx}\,dx=\int 3x^2\,dx \implies y+C_1=x^3+C_2 \implies \color{red}\boxed{y=x^3+C}$ . Note that we can combine the two constants into a new constant C.

Also, we may encounter differential equations with given conditions. These types of differential equations are called initial value problems (IVP). When solving a DE without conditions, we always find the General Solution to the DE. When an initial condition is applied, then we are finding a Particular Solution. Let's go through a quick example.

Example 2:

Solve $\frac{dy}{dx}=xe^{-x}$ ; $y(0)=0$ .

Directly integrate the DE:

$\int\frac{dy}{dx}\,dx=\int xe^{-x}\,dx$ .

We need to apply integration by parts to the integral on the right side.

$\int xe^{-x}\,dx$
let $u=x$ and $dv=e^{-x}$ . $\therefore \,du=\,dx$ and $v=-e^{-x}$ .

$\therefore \int xe^{-x}\,dx=-xe^{-x}+\int e^{-x}\,dx=-xe^{-x}-e^{-x}+C=-e^{-x}(x+1)+C$ .

Now apply the initial condition $y(0)=1$ .

$1=-e^{0}(0+1)+C \implies C=2$ .

Thus, our particular solution is : $\color{red}\boxed{y=-e^{-x}(x+1)+2}$ .

-------------------------------------------------------------------------

2. Separation of Variables

Another technique in solving differential equations is separation of variables. As the name suggests, we "separate" one variable from another in order to find a solution. Some of these are very straight forward, whereas some of the DE's require some thought. I will go through an easier example, and then a harder one:

Example 3:

Solve $\frac{dy}{dx}=4x^3y-y$ ; $y(1)=-3$ .

First, factor out a y and then separate the variables.

$\frac{dy}{dx}={\left(4x^3-1\right)}y \implies \frac{dy}{y}={\left(4x^3-1\right)}\,dx$ .

Integrate both sides and solve for y.

$\int\frac{dy}{y}=\int{\left(4x^3-1\right)}\,dx \implies ln{\left|y\right|}=x^4-x+C \implies {\left|y\right|}=e^{C}e^{x^4-x} \implies y=Ce^{x^4-x}$ .

Apply the initial condition $y(1)=-3$ .

$-3=Ce^{0} \implies C=-3$ .

$\therefore \color{red}\boxed{y=-3e^{x^4-x}}$ .

Example 4:

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

Factor the right hand side of the equation.

$x^2\frac{dy}{dx}=1-x^2+(1-x^2)y^2 \implies x^2\frac{dy}{dx}=(1-x^2)(1+y^2)$ .

Separate the variables and integrate.

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

$\int \frac{dy}{1+y^2}= \int\frac{1-x^2}{x^2}\,dx$
$tan^{-1}y=-\frac{1}{x}-x+C$ .

Solve for y.

$\color{red}\boxed{y=tan{\left(-\frac{1}{x}-x+C\right)}}$ .

-------------------------------------------------------------------------

I will post more later on today...after I sleep

The Following 22 Users Say Thank You to Chris L T521 For This Useful Post:
aleph1, arbolis, Chop Suey, Gaz, ILoveMaths07, janvdl, Johnaloa, lllll, Mathstud28, MAX09, o_O, Prove It, Rebesques, Schniz2, scorpion007, simplependulum, songoku, Stroodle, tasukete, wingless, Zero266, zxcv
$Donate to MHF$