Homogeneous linear second order differential equations

Homogeneous linear second order differential equations

Lessons

A Linear Second Order Differential Equation is of the form:

a(x)d2ydx2+b(x)dydx+c(x)y=d(x)a(x) \frac{d^2 y}{dx^2} +b(x) \frac{dy}{dx}+c(x)y=d(x)

Or equivalently,

a(x)y+b(x)y+c(x)y=d(x)a(x) y''+b(x) y'+c(x)y=d(x)

Where all of a(x),b(x),c(x),d(x)a(x),b(x),c(x),d(x) are functions of xx.

A Linear Second Order Differential Equation is called homogeneous if d(x)=0d(x)=0. So,

a(x)y+b(x)y+c(x)y=0a(x) y''+b(x) y'+c(x)y=0

And a general constant coefficient linear homogeneous, second order differential equation looks like this:

Ay+By+Cy=0Ay''+By'+Cy=0

Let’s suppose that both f(x) and g(x) are solutions to the above differential equations, then so is

y(x)=c1f(x)+c2g(x)y(x)=c_1 f(x)+c_2 g(x)

Where c1c_1 and c2c_2 are constants

Characteristic Equation
The general solution to the differential equation:

Ay+By+Cy=0Ay''+By'+Cy=0

Will be of the form: y(x)=erxy(x)=e^{rx}
Taking the derivatives:
y(x)=rerxy' (x)=re^{rx}
y(x)=r2erxy'' (x)=r^2 e^{rx}

And inputting them into the above equation:

Ar2erx+Brerx+Cerx=0Ar^2 e^{rx}+Bre^{rx}+Ce^{rx}=0

So we will have: erx(Ar2+Br+C)=0e^{rx} (Ar^2+Br+C)=0

The equation Ar2+Br+C=0Ar^2+Br+C=0 is called the Characteristic Equation. And is used to solve these sorts of questions.

Solving the quadratic we will get some values for rr:

Real Roots: r1r2r_1 \neq r_2
Complex Roots: r1,r2=λ±μir_1,r_2=\lambda \pm \mu i
Repeated Real Roots: r1=r2r_1=r_2

Though let’s deal with only real roots for now

As we will get two solutions to the characteristic equation:

y1(x)=er1xy_1 (x)=e^{r_1 x}
y2(x)=er2xy_2 (x)=e^{r_2 x}

So all solutions will be of the form:

y(x)=c1erx+c2erxy(x)=c_1 e^{rx}+c_2 e^{rx}
  • Introduction
    a)
    What are Homogeneous Linear Second Order Differential Equations? And what are some sorts of solutions to them?

    b)
    What is the Characteristic Equation?


  • 1.
    Solving Homogeneous Linear Second Order Differential Equations
    Find some general solutions to the following constant coefficient homogeneous linear second order differential equation:

    2y+5y3y=02y''+5y'-3y=0

  • 2.
    Using the initial conditions find a particular solution to the following differential equation:

    y+3y4y=0y''+3y'-4y=0
    y(0)=3,y(0)=2y(0)=3, y' (0)=-2