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

Answer 1

The General Solution to the DE # y'' - 4y' = 2x^2 # is

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

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 # y'' - 4y' = 0 #, the Axillary Equation is:
# m^2-4m+0=0# # :. m(m-4)=0#
This has two distinct real solutions, #m=0# and #m=4#

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 # y'' - 4y' = 2x^2 # we get:
# (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'' - 4y' = 2x^2 # is
# 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
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Answer 2

To find the general solution of the given differential equation (y'' - 4y' = 2x^2), you first solve the associated homogeneous equation (y'' - 4y' = 0), which has the characteristic equation (r^2 - 4r = 0). The roots of this equation are (r_1 = 0) and (r_2 = 4), yielding the homogeneous solution (y_h(x) = C_1 + C_2e^{4x}), where (C_1) and (C_2) are constants.

For the particular solution, since the right-hand side of the original equation is (2x^2), assume a particular solution of the form (y_p(x) = ax^2 + bx + c), where (a), (b), and (c) are constants to be determined.

Differentiate (y_p(x)) twice and substitute into the original differential equation. Then, equate coefficients of like terms to solve for (a), (b), and (c).

After finding (a), (b), and (c), the general solution is given by (y(x) = y_h(x) + y_p(x)), where (y_h(x)) is the homogeneous solution and (y_p(x)) is the particular solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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