Homogeneous linear second order differential equations

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  1. What are Homogeneous Linear Second Order Differential Equations? And what are some sorts of solutions to them?
  2. 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:

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

      y(0)=3,y(0)=2y(0)=3, y' (0)=-2
      Topic Notes
      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:


      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:


      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}