How do you find all solutions of the differential equation #(d^2y)/(dx^2)=3y#?

1 Answer
Dec 31, 2016

# \ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)#

Where #A,B# are arbitrary constants.

Explanation:

# (d^2y)/dx^2=3y => (d^2y)/dx^2 + 0dy/dx-3y=0#

This is a Second Order Homogeneous Differential Equation which we solve as follows:

We look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

# m^2+0m-3 = 0#
# :. m^2-3 = 0#
# :. m = +-sqrt(3)#

Because this has two distinct real solutions #sqrt(3)# and #-sqrt(3)#, the solution to the DE is;

# \ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)#

Where #A,B# are arbitrary constants.

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:

# \ \ \ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)#

# :. \ y' = Asqrt(3)e^(sqrt(3)x) - Bsqrt(3)e^(-sqrt(3)x) #

# :. y'' = Asqrt(3)sqrt(3)e^(sqrt(3)x) + Bsqrt(3)sqrt(3)e^(-sqrt(3)x) #
# :. y'' = 3Ae^(sqrt(3)x) + 3Be^(-sqrt(3)x) #

Hence;

# y'' - 3y = 3Ae^(sqrt(3)x) + 3Be^(-sqrt(3)x) -3{Asqrt(3)e^(sqrt(3)x) + Bsqrt(3)e^(-sqrt(3)x)} = 0#
-+-+-+-+-+-+-+-+-+-+-+-+-+-+