What is the general solution of the differential equation # (d^2y)/(dx^2) + 3dy/dx + 2y = 2sinx#?

1 Answer
Jul 21, 2017

# y(x) = Ae^(-x)+Be^(-2x) -3/5cosx+1/5sinx #

Explanation:

We have:

# (d^2y)/(dx^2) + 3dy/dx + 2y = 2sinx# ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

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

And it's associated Auxiliary equation is:

# m^2+3m+2 = 0 #
# (m+1)(m+2) = 0 #

Which has two real and distinct solutions #m=-1,-2#

Thus the solution of the homogeneous equation is:

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

Particular Solution

With this particular equation [A], a probably solution is of the form:

# y = acos(x)+bsin(x) #

Where #a# and #b# are constants to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \= -asin(x)+bcos(x) #
# y'' = -acos(x)-bsin(x) #

Substituting into the initial Differential Equation #[A]# we get:

# -acos(x)-bsin(x) + 3{-asin(x)+bcos(x)} + 2{acos(x)+bsin(x)}=2sin(x)#

# -acos(x)-bsin(x) -3asin(x)+3bcos(x) + 2acos(x)+2bsin(x)=2sin(x)#

Equating coefficients of #cos(x)# and #sin(x)# we get:

#cos(x): -a+3b+2a=0 => a+3b=0#
#sin(x): -b -3a+2b=2 \ => b-3a=2 #

Solving simultaneously we get:

# a=-3/5 # and #b=1/5#

And so we form the Particular solution:

# y_p = -3/5cosx+1/5sinx#

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^(-x)+Be^(-2x) -3/5cosx+1/5sinx #