What is the general solution of the differential equation? : # 2y'' + 3y' -y =0#

1 Answer
Jun 10, 2017

# y = Ae^((-3/4-sqrt(17)/4)x)+Be^((-3/4+sqrt(17)/4)x) #

Explanation:

We have:

# 2y'' + 3y' -y =0#

This is a second order linear Homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

# 2m^2+3m-1=0 #

This has two distinct real solutions:

#m_1=-3/4-sqrt(17)/4# and #m_2=-3/4+sqrt(17)/4#

And so the solution to the DE is;

# \ \ \ \ \ y = Ae^(m_1x)+Be^(m_2x) # Where #A,B# are arbitrary constants
# :. y = Ae^((-3/4-sqrt(17)/4)x)+Be^((-3/4+sqrt(17)/4)x) #

Note

The given solution:

# y=y_1=e^(-2x) #

is not actually a solution of the given DE, as:

# y' \ \ =-2e^(-2x) #
# y''=4e^(-2x) #

And

# 2y'' + 3y' -y =8e^(-2x) -6e^(-2x) -e^(-2x) ne 0#