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

##### 1 Answer

# 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,

**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

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

Let us assume the above solution works, in which case be differentiating wrt

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

# y'' = -acos(x)-bsin(x) #

Substituting into the initial Differential Equation

# -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): -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 #