How to solve second-order homogeneous linear differential equations w/ constant coefficients: basic idea

The tutor shows the quadratic method (pka, characteristic equation) behind 2nd order LDEs w/ constant coefficients.

An exponential function of the form

y=ekt

has the first derivative

y’=kekt

and second derivative

y”=k2ekt

Seeing an equation like, for example,

y” -3y’ +2y = 0

one can suppose the solution to be of the form

y=Cekt (C is some constant)

so the equation becomes

k2Cekt -3kCekt +2Cekt = 0

Now we can factor out Cekt:

Cekt(k2 -3k +2)=0

and then factor and solve for k:

Cekt(k-2)(k-1)=0

k=2 or k=1

Therefore, a possible solution to the equation

y” -3y’ +2y = 0

is

y=C1e1t + C2e2t

or just

y=C1et + C2e2t

HTH:)

Source:

Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems. New York: John Wiley & Sons, Inc., 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Differential equations: exact differential equation

The tutor explores how to detect and solve exact differential equations with a very simple example.

When a differential equation of the form

P(x,y) +Q(x,y)y’ = 0

results from the implicit differentiation of an original equation F(x,y)=c, the equation

P(x,y) +Q(x,y)y’ = 0

is said to be an exact differential equation.

The way to tell is that, for an exact differential equation,

Py(x,y) = Qx(x,y)

Example: Solve the differential equation

siny +1 +xy’cosy +2y’ = 0

First, we render it to P(x,y) +Q(x,y)y’ = 0:

siny + 1 +(xcosy +2)y’ = 0

P(x,y) = siny+1; Q(x,y) = xcosy + 2

Now we take the “other derivative” of each one:

Py(x,y) = cosy

Qx(x,y) = cosy

Py(x,y) = Qx(x,y): the equation is exact. Therefore,

siny +1 +xy’cosy +2y’ = 0

is the implicit derivative of some equation F(x,y) = c, which we need to find. Furthermore,

siny + 1 = Fx

We integrate siny + 1 with respect to x:

∫siny +1 = xsiny + x + g(y)

where g(y) is a function only of y that was lost in the original derivative by x.

g'(y) should be recognizable in Q(x,y): specifically, it’s 2⇒ g(y)=2y

Therefore, the solution F(x)=c is

xsiny +x +2y=c

Note that its implicit derivative is

siny + 1 + x(cosy)y’ +2y’ = 0

which matches the original differential equation posed.

Source:

Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems. Toronto: John Wiley & Sons, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.