## - 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
$$\displaystyle y=Ae^{-{b\over a}x}$$

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 $$y=Ae^{αx}+Be^{βx}$$
$y=Ae^x+Be^{4x}$

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 $$y=e^{αx}(Ax+B)$$
$$\displaystyle y=e^{{1\over2} x}(Ax+B)$$

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 $$y=e^{αx} (A \cos ⁡βx+B \sin ⁡βx)$$
$$y=e^x(A \cos ⁡2x+B \sin ⁡2x)$$