Solve the differential equation: #(d^2y)/ (dx^2) −8 (dy)/(dx) =−16y#? Discuss what kind of differential equation is this, and when it may arise?

2 Answers
Jul 18, 2016

#y = (Ax + B) e^(4x)#

Explanation:

#(d^2y)/ (dx^2) −8 (dy)/(dx) =−16y#

best written as

#(d^2y)/ (dx^2) −8 (dy)/(dx) + 16y = 0 qquad triangle#

which shows that this is linear second order homogeneous differential equation

it has characteristic equation

#r^2 −8 r + 16 = 0#

which can be solve as follows

#(r-4)^2 = 0, r = 4#

this is a repeated root so the general solution is in form

#y = (Ax + B) e^(4x)#

this is non-oscillating and models some kind of exponential behaviour that really depends on the value of A and B. One might guess it could be an attempt to model population or predator/prey interaction but i can't really say anything very specific.

it shows instability and that's about all i could really say about it

Jul 18, 2016

#y=(C_1+C_2x)e^{lambda x}#

Explanation:

The differential equation

#(d^2y)/(dx^2)-8(dy)/(dx)+16y=0#

is a linear homogeneous constant coefficient equation.

For those equations the general solution has the structure

#y = e^{lambda x}#

Substituting we have

#e^{lambda x}(lambda^2-8lambda+16) = 0#

Here #e^{lambda x} ne 0# so the solutions must obey

#lambda^2-8lambda+16 = (lambda-4)^2= 0#

Solving we obtain

#lambda_1=lambda_2=4#

When the roots repeat, #d/(d lambda)e^{lambda x}# is also solution. In case of #n# roots repeated, we will have as solutions:

#C_i (d^i)/(d lambda ^i) e^{lambda x}# for #i=1,2,cdots,n#

So, to maintain the number of initial conditions, we include them as independent solutions.

In this case we have

#y = C_1 e^{lambda x}+C_2d/(d lambda)e^{lambda x}#

which results in

#y=(C_1+C_2x)e^{lambda x}#

Those equations appear when modelling linear lumped parameter systems like those found in linear circuit theory or linear mechanics. Those equations are normally handled using operational algebraic methods like Laplace Transform methods
https://en.wikipedia.org/wiki/Laplace_transform