Question

What is the general solution of the differential equation `y''' - y'' = e^xcosx`?

Answer 1

`y = -1/2e^xcos(x)+ c_1e^x+ c_2x + c_3`

Explanation:

Given: `y'''-y''=e^x cos(x)" [1]"`

Let u = `y''`, then `u' = y'''` and substitute into equation [1]:

`u' - u = e^xcos(x)" [2]"`

The integrating factor is `I = e^(int-1dx) = e^-x`

Multiply both sides of equation two by `I`:

`e^-xu' - e^-xu = e^-xe^xcos(x)`

`e^-xu' - e^-xu = cos(x)`

We know that the left side integrates to the product of `Iu` and the right side is well known:

`e^-xu = sin(x)+c_1`

`u = e^x(sin(x)+ c_1)`

Reverse the substitution:

`y'' = e^x(sin(x)+ c_1)`

`y' = inte^xsin(x)dx+ c_1inte^xdx`

`y' = 1/2e^x(sin(x)-cos(x))+ c_1e^xdx+ c_2`

`y = 1/2inte^xsin(x)dx-1/2inte^xcos(x)+ c_1inte^xdx + c_2intdx`

`y = 1/4e^x(sin(x)-cos(x))-1/4e^x(sin(x)+cos(x) + c_1e^x+ c_2x+ c_3`

`y = -1/2e^xcos(x)+ c_1e^x+ c_2x + c_3`

Answer 2

`y(x) = Ax+B + Ce^x -1/2e^xcosx`

Explanation:

We have:

`y''' - y'' = e^xcosx` ..... [A]

This is a Third order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complimentary Function

The homogeneous equation associated with [A] is

`y''' -y'' = 0`

And it's associated Auxiliary equation is:

`m^3 -m^2 = 0`
`m^2(m-1) = 0`

Which has repeated solutions `m=0`, and a real distinct solution `m=1`

Thus the solution of the homogeneous equation is:

`y_c = (Ax+B)e^(0x) + Ce^(1x)`
`\ \ \ = Ax+B + Ce^x`

Note this solution has `3` constants of integration and is therefore the complete solution for the homogeneous equation.

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

`y = e^x(acosx + bsinx)`

Where the constants `a` and `b` are to be determined by direct substitution and comparison:

Differentiating wrt `x` (using the product rule) we get:

`y' = e^x(-asinx + bcosx) + e^x(acosx + bsinx)`
`\ \ \ = e^x(acosx -asinx + bcosx + bsinx)`

Differentiating again wrt `x` (using the product rule) we get:

`y'' = e^x(-asinx -acosx - bsinx + bcosx) + e^x(acosx -asinx + bcosx + bsinx)`
`\ \ \ \ = e^x(-2asinx + 2bcosx)`

Differentiating again wrt `x` (using the product rule) we get:

`y''' = e^x(-2acosx - 2bsinx) + e^x(-2asinx + 2bcosx)`
`\ \ \ \ = e^x(-2acosx - 2bsinx -2asinx + 2bcosx)`
`\ \ \ \ = e^x(-2acosx -2asinx - 2bsinx + 2bcosx)`

Substituting into the DE [A] we get:

`y''' - y'' = e^xcosx`

`:. e^x(-2acosx -2asinx - 2bsinx + 2bcosx) - e^x(-2asinx + 2bcosx) = e^xcosx`

`:. -2acosx -2asinx - 2bsinx + 2bcosx - +2asinx - 2bcosx = cosx`

Equating coefficients of `cosx` and `sinx` we get

`cos: \ \ -2a + 2b - 2b = 1 => -2a=1 => a=-1/2`
`sin: \ \ -2a - 2b + 2a = 0 => -2b=0 => b =0`

And so we form the Particular solution:

`y_p = e^x(-1/2cosx + 0)`
`\ \ \ = -1/2e^xcosx`

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Ax+B + Ce^x -1/2e^xcosx`