# What is the general solution of the differential equation? : # (d^2y)/dx^2-dy/dx-2y=4x^2 #

##### 1 Answer

# y = Ae^(-x)+Be^(2x) -2x^2+2x-3#

#### Explanation:

There are two major steps to solving Second Order DE's of this form:

# (d^2y)/dx^2-dy/dx-2y=4x^2 #

**1) Find the Complementary Function (CF)**

This means find the general solution of the Homogeneous Equation

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

To do this we look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

#m^2-m-2=0#

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

This has two distinct real solutions,

And so the solution to the DE is;

# y = Ae^(-x)+Be^(2x) # Where#A,B# are arbitrary constants

-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Verification:

If

# y = Ae^(-x)+Be^(2x) # , then

# y' = -Ae^(-x)+2Be^(2x) #

# y'' = Ae^(-x)+4Be^(2x) # And so,

# y''-y'-2y = Ae^(-x)+4Be^(2x) -(-Ae^(-x)+2Be^(2x))-2(Ae^(-x)+Be^(2x)) = 0#

-+-+-+-+-+-+-+-+-+-+-+-+-+-+

**2) Find a Particular Integral* (PI)**

This means we need to find a specific solution (that is not already part of the solution to the Homogeneous Equation). As the RHS is a quadratic we try a solution of the quadratic form:

# y=ax^2+bx+c #

Where

If

# y'\ = 2ax+b #

# y'' = 2a #

If we substitute into the initial DE we get:

# \ \ 2a - (2ax+b) - 2(ax^2+bx+c)=4x^2 #

# :. 2a - 2ax-b - 2ax^2-2bx-2c = 4x^2 #

Equating Coefficients we have

#Coef(x^2) : -2a=4 \ \ \ \ \ \ \ \ => a=-2#

#Coef(x^1) : -2a-2b=0 => b=2#

#Coef(x^0) : 2a-b-2c=0 \ \ \ \ => c=-3 #

So we have found that a Particular Solution is:

# y = -2x^2+2x-3 #

**3) General Solution (GS)**

The General Solution to the DE is then:

GS = CF + PI

Hence The General Solution to the initial DE is

# y = Ae^(-x)+Be^(2x) -2x^2+2x-3#