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
