How to Solve Second Order Differential Equations

a brief over view of homo-genius second order differential equation and the three most common forms Standard Roots,Double Roots, and Imaginary Roots

Things You'll Need

• pencil
• paper
• calculator (optional)

Instructions

1. Given a homogeneous second order differential equation of the form ay^''+by^'+cy=0 find the general solution.

   Find the characteristic equation:
   The characteristic equation will always have the form
   on ar^2+br+c=0.
   A, B, C are arbitrary constants and are the same as those in
   your original Diff equation.

   If A≠1 then divide by A to get the equation in its standard form.

   If the original differential equation ≠0 then that means the method of Chain Rule must be used.

2. factor the Characteristic equation to find it roots.
   • Use factoring or the quadratic equation to find the roots.
   • The root will take 1 of 3 possible forms, standard, Double Roots, or imaginary this will determine the form of your solution

3. find the General solution using the roots of the characteristic equation

   Standard roots: give the form y=C_1 e^(tr_1 )+C_2 e^(〖tr〗_2 )
   Where C_1,C_2 = arbitrary constants r_1,r_2 =
   the character roots

   Double roots: give the form y=C_1 e^(tr_1 )+C_2 te^(〖tr〗_2 )
   This is when r_1=r_2 so the t is added to give
   an independent second solution.
   Imaginary root: give the form y=C_1cos(μt) e^λt+C_2 sin(μt) e^λt
   Where μ ,λ are given by the roots that take the
   form r=λ+iμ, r=λ+iμ
   This comes from the quadratic
   equation r=(-b±√(b^2-4ac))/2a

4. find the values for C_1,C_2
   This step requires Initial condition or (IC) these values are

   given in the problem statement
   IC. y(t=0)=0, y' (t=0)=1 note: the number does not have to be zero and can be
   any real number. The IC will be given in the problem statement.

   Using the equation found in the last step y=C1 e^(tr1 )+C2 e^(tr2 ) differentiate to find
   y^'=C_1 r_1 e^(tr_1 )+C_2 〖r_2 e〗^(〖tr〗_2 )

   Then solve for C_1,C_2 by plugging in the IC.

   This gives you two equation and two unknowns solve.

   Note: if t=0 then you get e^0=1

5. final answers

   After C_1,C_2 are found plug the back into the original equation

   y=C_1 e^(tr_1 )+C_2 e^(〖tr〗_2 )
   Where C_1,C_2 , r_1,r_2 should all be real numbers.

   examples problem:
   Standard roots: y^''+2y^'+3y=0 , let I.C = y(0)=0, y^' (0)=1
   1. Char equ. Is r^2+2r+3=0 ⟹ by factoring we get
   (r-1)(r+3)⟹ r= 1 ,r= -3
   2. Gen solution: y=C_1 e^(1(t))+C_2 e^(-3t)

   3. find y' : y^'=1C_1 e^1(t) -3C_2 e^(-3t)

   4. solve for C_1,C_2 by using I.C:
   y(0)=0=C_1 e^1(0) +C_2 e^(-3(0) )
   ⟹ 0=C_1+C_2
   y^' (0)=1=1C_1 e^1(0) -3C_2 e^(-3(0) )
   ⟹1=C_1-3C_2
   Solving gives: C_1= -1/2, C_2=1/2 plug in for final answer is:

   y=〖-1/2 e〗^t+〖1/2 C〗_2 e^(-3t)