# If a particular integral of the differential equation #(D^2+2D-1)y=e^(ax)# is #(-4/7)e^(ax)# then the value of a is ?

##### 1 Answer

# a=-3/2, -1/2# #

We cannot eliminate a solution, thus we are left with two possibilities

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-3/2x) #

or

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-1/2x) #

#### Explanation:

We have:

# (D^2+2D-1)y = e^(ax) # with a PI,#-4/7e^(ax) #

Or, In standard form:

# y'' + 2y' -y = e^(ax) #

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution,

**Complementary Function**

The homogeneous equation associated with [A] is

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

And it's associated Auxiliary equation is:

# m^2+2m-1 = 0 => (m+1)^2-1-1 = 0#

Which has two real and distinct solution

Thus the solution of the homogeneous equation is:

# y_c = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) #

**Particular Solution**

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

# y = Ke^(ax) #

Where the constants

Differentiating wrt

# \ y' = aKe^(ax) #

# y'' = a^2Ke^(ax) #

Substituting into the DE [A] we get:

# (a^2Ke^(ax)) + 2(aKe^(ax)) - (Ke^(ax)) = e^(ax) #

# :. K(a^2 + 2a - 1) = 1 #

# :. K = 1/(a^2 + 2a - 1) #

We are also given that the PS is

# -4/7e^(ax) = Ke^(ax) => K=-4/7#

Allowing us to find

# 1/(a^2 + 2a - 1) = -4/7 #

# :. 4(a^2 + 2a - 1) = -7 #

# :. 4a^2 + 8a +3=0 #

# :. (2a+1)(2a+3) = 0 #

# :. a=-3/2, -1/2# #

We cannot eliminate a solution, thus we are left with two possibilities, Allowing us to write the General Solution

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-3/2x) #

or

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-1/2x) #