# How do you find the general solution of the differential equations #y''-4y'=2x^2#?

##### 1 Answer

The General Solution to the DE

# y = A + Be^(4x) -1/6x^3 - 1/8x^2 - 1/16x #

#### Explanation:

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

# y'' - 4y' = 2x^2 #

**Find the Complementary Function (CF)**

This means find the general solution of the Homogeneous Equation

# y'' - 4y' = 0 #

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

As

# m^2-4m+0=0#

# :. m(m-4)=0#

This has two distinct real solutions,

And so the solution to the DE is;

# \ \ \ \ \ y = Ae^(0x) + Be^(4x)# Where#A,B# are arbitrary constants

# :. y = A + Be^(4x) #

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

Verification:

# y = A + Be^(4x) => y' = Be^(4x) #

# " "\ => y'' = 16Be^(4x) #

# :. y'' - 4y' = 16Be^(4x) -4(4Be^(4x)) #

# " "\ = 16Be^(-4x) -16Be^(-4x) #

# " "\ = 0 #

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

**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).

We look for a particular solution which is a combination of functions on the RHS of the form

# y = ax^3 + bx^2 + cx + d #

In which case, we get:

# y' \ = 3ax^2 + 2bx + c #

# y' '= 6ax + 2b #

If we substitute into the DE

# (6ax + 2b) - 4(3ax^2 + 2bx + c) = 2x^2 #

Equating Coefficients we get:

#x^2: -12a=2 => a=-1/6#

#x^1: 6a-8b=0 => b=-1/8#

#x^0: 2b-4c=0 => c=-1/16#

So we have found that a Particular Solution is:

# y = -1/6x^3 - 1/8x^2 - 1/16x + d #

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

Verification:

# y = -1/6x^3 - 1/8x^2 - 1/16x + d #

# " " => y'' = -1/2x^2-1/4x-1/16 #

# " " => y'' = -x-1/4 #

# :. y'' - 4y' = (-x-1/4) -4(-1/2x^2-1/4x-1/16) #

# " "\ = -x-1/4+2x^2+x+1/4 #

# " "\ = 2x^2 #

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

**General Solution (GS)**

The General Solution to the DE is then:

GS = CF + PI

Hence The General Solution to the DE

# y = A + Be^(4x) -1/6x^3 - 1/8x^2 - 1/16x + d#

# y = A + Be^(4x) -1/6x^3 - 1/8x^2 - 1/16x # where A is arbitary