- Homogeneous Differential Equations

\(ay'+by=0\) and \(ay''+by'+cy=0\)
are first and second order homogenous equations where \(a, b\) and \(c\) are constants. The solution of these equations is called the complementary function (C.F).
LESSON 1
Solve the differential equation
\(\displaystyle 5 {dy\over dx}-2y=0\)
SOLUTION
  • Rearrange and separate variables
\(\displaystyle 5 {dy\over dx}=2y\)
\(\displaystyle {dy\over dx}={2\over5} y\)
\(\displaystyle {1\over y}\:dy={2\over5}\: dx\)
  • Integrate both sides with respect to the required variable.
\(\displaystyle ∫{1\over y}\:dy=∫{2\over5}\: dx\)
\(\displaystyle \ln ⁡y={2\over5}x+c\)
  • Simplify by taking e of both sides.
\(\displaystyle e^{\ln ⁡y}=e^{{2\over5}x+c}\)
\(\displaystyle y=e^c e^{{2\over5}x}\)
\(\displaystyle y=Ae^{{2\over5}x}\)
In general, the solution (complementary function) of a first order differential equation of the form
\(\displaystyle a {dy\over dx}+by=0\)
is
\(\displaystyle y=Ae^{-{b\over a}x}\)

VIDEO LESSON



Auxiliary Quadratic Equation
Given the equation
\(\displaystyle a{d^2y\over dx^2}+b {dy\over dx}+cy=0\)
or
\(ay''+by'+cy=0\)
The complementary function is determined by the roots and the nature of the roots of the quadratic auxiliary equation which is
\(au^2+bu+c=0\)

Nature of roots of auxiliary equation

\(au^2+bu+c\)

Complementary function

Real and distinct, \(\alpha\) and \(\beta\)

\(y=Ae^{\alpha x}+Be^{\beta x}\)

Repeated \(\alpha\)

\(y=e^{\alpha x}(Ax+B)\)

Complex roots, \(\alpha\pm \beta i\)

\(y=e^{\alpha x}(A\cos \beta x+Bi\sin \beta x)\)

LESSON 2
Solve the equation
\(y''-5y'+4y=0\)
SOLUTION
\(y''-5y'+4y=0\)
  • Determine the auxiliary equation
Auxiliary equation
\(u^2-5u+4=0\)
  • Determine the roots of the auxiliary equation.
\(u=1, 4\)
The roots are real and distinct so we substitute the roots into the equation \(y=Ae^{αx}+Be^{βx}\)
\[y=Ae^x+Be^{4x}\]

VIDEO LESSON


LESSON 3
Solve the equation
\(4y''-4y'+y=0\)
SOLUTION
\(4y''-4y'+y=0\)
  • Write the auxiliary equation.
Auxiliary equation
\(4u^2-4u+1=0\)
  • Determine the roots of the auxiliary equation
\(\displaystyle u={1\over2}\) (twice)
The auxiliary equation has a repeated root therefore substitute this value into the equation \(y=e^{αx}(Ax+B)\)
\(\displaystyle y=e^{{1\over2} x}(Ax+B)\)

VIDEO LESSON


LESSON 4
Solve the equation
\(y''-2y'+5y=0\)
SOLUTION
\(y''+2y'+5y=0\)
  • State the auxiliary equation
Auxiliary equation
\(u^2+2u+5=0\)
  • Determine the roots of the auxiliary equation.
\(u=1\pm 2i\)
The auxiliary equation has complex roots therefore substitute the real and imaginary parts into the equation \(y=e^{αx} (A \cos ⁡βx+B \sin ⁡βx)\)
\(y=e^x(A \cos ⁡2x+B \sin ⁡2x)\)

VIDEO LESSON