Solve the differential equation:

d2y/dx2 - 2 dy/dx + y = 3sinhx

The answer should be:
y(x) = e^x (Ax+ B) + ( 3^8 )(2[x^2][e^x] - [e^-x] )

Can someone please show me how to work it out?

There is probably a mistake in the transcription of the solution:

y(x) = e^x (Ax+ B) + ( 3/8 )(2[x^2][e^x] - [e^-x] )
Namely the power 3^8 should be a division as shown in bold.

Given:
y"-2y'+y=3sinh(x) .....(0)

This is a linear (y and its derivatives appear in first power) second order (maximum y") non-homogeneous (right-hand side is not zero) differential equation with constant coefficients (coefficients of y and its derivatives are constants).

The general solution of the homogeneous equation with constant coefficients is a combination of terms Aixjek where i,j,k could be complex.

We will first solve the homogeneous equation, namely drop the right hand side to get:
y"-2y'+y=0 .....(1)
From (1), we form and solve the characteristic polynomial in z which has the same coefficients as (1).
z²-2z+1=0......(2)
to get
(z-1)(z-1)=0
or z=1 (multiplicity 2).
The general solution of the homogeneous equation is a linear combination of the basis formed by the solutions of the characteristic equation (2):
Aiezi.
This applies when the solutions of zi are distinct. If multiplicity occurs, as in this case, terms will contain successive powers of x.
Therefore with a multiplicity of 2 for z=1, the solution to the homogeneous equation is:
k1xex+k2ex
where the power of e is 1, equivalent to the solution z=1.
To determine the solution of the non-homogeneous equation, we will find a particular solution.

Since the right hand side is 3sinh(x)=(3/2)(ex-e-x),
we can use the method of undetermined coefficients.
Since one of the terms on the right-hand-side (ex) corresponds to one of the terms in the solution basis, we need to multiply the term by powers of x until it is distinct from any term of the general solution.
The assumed particular will then take the form:
yp(x) = Ax²ex+Be-x

The next step is to find the coefficients of A and B.
This can be done by substituting y(p) into equation (0), carry out the differentiations and compare coefficients of ex and e-x to get
A=3/4, and B=-3/8

The solution to the differential equation (0) is therefore the sum of the general solution and the particular solution:

y=(3/4)x²ex - (3/8)e-x + (k1x+k2)ex

where k1 and k2 are integration constants.

The above expression can be rearranged to be identical with the given answer.

References for further reading:
Differential Equations by Richard Bronson & Gabriel Costa, Shaum's Outline Series, McGraw-Hill.
http://en.wikipedia.org/wiki/Linear_differential_equation
http://en.wikipedia.org/wiki/Differential_equation

what is the answer to this problem

6(6g-2)+8(1-5g)=2g

To solve the given differential equation, we can use the method of undetermined coefficients. This method involves guessing a particular solution based on the right-hand side of the equation and finding a complementary solution to satisfy the homogeneous equation.

Let's break down the steps:

Step 1: Homogeneous Solution
First, we find the solution to the homogeneous equation, i.e., when the right-hand side is zero.

The homogeneous equation is:
d^2y/dx^2 - 2(dy/dx) + y = 0

To solve this second-order linear homogeneous equation, we assume a solution of the form:
y_h(x) = e^(rx)

Taking the derivative of y_h(x), we have:
y_h'(x) = re^(rx)
y_h''(x) = r^2e^(rx)

Substituting these into the homogeneous equation, we get:
r^2e^(rx) - 2re^(rx) + e^(rx) = 0

Factoring out e^(rx), we have:
e^(rx)(r^2 - 2r + 1) = 0

The characteristic equation is:
r^2 - 2r + 1 = 0

Solving this quadratic equation, we find that r = 1 (with multiplicity 2).

Therefore, the homogeneous solution is:
y_h(x) = c1e^x + c2xe^x, where c1 and c2 are arbitrary constants.

Step 2: Particular Solution
Next, we need to find a particular solution y_p(x) that satisfies the non-homogeneous equation when the right-hand side is 3sinh(x).

Since the right-hand side is of the form A*sinh(x), we guess a solution of the form:
y_p(x) = B*x^2*e^x

Taking the derivatives of y_p(x), we have:
y_p'(x) = (2Bx + Bx^2)e^x
y_p''(x) = (2B + 4Bx + Bx^2)e^x

Substituting these into the non-homogeneous equation, we get:
(2B + 4Bx + Bx^2)e^x - 2(2Bx + Bx^2)e^x + Bx^2e^x = 3sinh(x)

Simplifying, we have:
(2B + 4Bx + Bx^2 - 4Bx - 2Bx^2 + Bx^2)e^x = 3sinh(x)
(B + Bx^2)e^x = 3sinh(x)

Comparing the coefficients of sinh(x), we find:
B + Bx^2 = 3

Solving this equation, we have B = 3.

Therefore, the particular solution is:
y_p(x) = 3x^2e^x

Step 3: Complete Solution
The complete solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:
y(x) = y_h(x) + y_p(x)
= c1e^x + c2xe^x + 3x^2e^x

Finally, we can simplify the equation:
y(x) = e^x(c1 + c2x) + 3x^2e^x

Comparing this solution to the given answer:
y(x) = e^x(Ax + B) + (3^8)(2[x^2][e^x] - [e^-x])

We can see that A = c2 and B = c1 for the general solution. Additionally, the term (3^8)(2[x^2][e^x] - [e^-x]) can be rearranged to match 3x^2e^x, given that 3^8 = 1.

Hence, the given answer: y(x) = e^x(Ax + B) + (3^8)(2[x^2][e^x] - [e^-x])