Question

Can you please solve this in calculus? Thanks `y''''+2y'''+2y''+2y'+y=1/2cosx+xe^x`

Answer 1

`y = Axe^(-x) + Be^(-x) + Ccosx+Dsinx -1/8xcosx+1/8xe^x-1/4e^x`

Explanation:

We have:

`y''''+2y'''+2y''+2y'+y=1/2cosx+xe^x` ..... [A]

This is a fourth 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.

Complementary Function

The homogeneous equation associated with [A] is

`y''''+2y'''+2y''+2y'+y = 0` ..... [B]

And it's associated Auxiliary equation is:

`m^4+2m^3+2m^2+2m+1 = 0`

The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. By consideration of a graph of the function, we find that `m=-1` is a repeated root, so:

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

And so we have the solutions:

`m=-1 \ \ \ \` (reopeated)
`m=+-i`

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

• Real distinct roots `m=alpha,beta, ...` will yield linearly independent solutions of the form `y_1=Ae^(alphax)`, `y_2=Be^(betax)`, ...
• Real repeated roots `m=alpha`, will yield a solution of the form `y=(Ax+B)e^(alphax)` where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) `m=p+-qi` will yield a pairs linearly independent solutions of the form `y=e^(px)(Acos(qx)+Bsin(qx))`

Thus the solution of the homogeneous equation [B] is:

`y = (Ax+B)e^(-x) + e^(0x)(Ccosx+Dsinx)`
`\ \ = Axe^(-x) + Be^(-x) + Ccosx+Dsinx`

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

`y'''-y''+4y' = f(x) \ \` with `f(x)=1/2cosx+xe^x`

We would look for a solution of the form:

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

However, as a form of `acosx+bsinx` already exists in the CF solution and so we must consider a potential solution of the form:

`y = axcosx+bxsinx+(cx+d)e^x` ..... [C]

Where the constants `a,b,c,d` is to be determined by direct substitution and comparison:

Differentiating [C] wrt `x` four times we get:

`y^((1)) = (ax)(-sinx) + (a)(cosx) + (bx)(cosx) + (b)(sinx)`
`\ \ \ \ \ \ \ = + (cx+d)(e^x) + (c)(e^x)`

`\ \ \ \ \ \ \ = (a+bx)cosx + (b-ax)sinx+(cx+c+d)e^x`

`y^((2)) = (a+bx)(-sinx) + (b)(cosx) + (b-ax)(cosx) + (-a)(sinx)`
`\ \ \ \ \ \ \ = + (cx+c+d)(e^x) + (c)(e^x)`

`\ \ \ \ \ \ \ = (2b-ax)cosx + (-2a-bx)sinx+(cx+2c+d)e^x`

`y^((3)) = (2b-ax)(-sinx) + (-a)(cosx) + (-2a-bx)(cosx) + (-b)(sinx)`
`\ \ \ \ \ \ \ = + (cx+2c+d)(e^x) + (c)(e^x)`

`\ \ \ \ \ \ \ = (-3a-bx)cosx + (-3b+ax)sinx+(cx+3c+d)e^x`

`y^((4)) = (-3a-bx)(-sinx) + (-b)(cosx) + (-3b+ax)(cosx) + (a)(sinx)`
`\ \ \ \ \ \ \ = + (cx+3c+d)(e^x) + (c)(e^x)`

`\ \ \ \ \ \ \ = (-4b+ax)cosx + (4a+bx)sinx+(cx+4c+d)e^x`

Substituting these results into the DE [A] we get:

`(-4b+ax)cosx + (4a+bx)sinx+(cx+4c+d)e^x`
`\ \ \ \ \ + 2{(-3a-bx)cosx + (-3b+ax)sinx+(cx+3c+d)e^x}`
`\ \ \ \ \ + 2{(2b-ax)cosx + (-2a-bx)sinx+(cx+2c+d)e^x}`
`\ \ \ \ \ + 2{(a+bx)cosx + (b-ax)sinx+(cx+c+d)e^x}`
`\ \ \ \ \ + {axcosx+bxsinx+(cx+d)e^x}`
`\ \ \ \ \ = 1/2cosx+xe^x`

Equating coefficients we get:

`sinx \ \ \ \ : 4a-6b-4a+2b = 0 => b=0`
`xsinx \ : b +2a-4b-2a+b = 0`

`cosx \ \ \ \ : -4b-6a+4b+2a = 1/2 => a = -1/8`
`xcosx \ : a-2b-2a+2b+a= 0`

`xe^x \ \ \ \ \ \ : c+ 2c + 2c + 2c + c = 1 => c=1/8`
`e^x \ \ \ \ \ \ \ \ : 4c+d+6c+2d+4c+2d+2c+2d+d = 0 => d=-1/4`

And so we form the Particular solution:

`y_p = -1/8xcosx+(1/8x-1/4)e^x`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Axe^(-x) + Be^(-x) + Ccosx+Dsinx -1/8xcosx+1/8xe^x-1/4e^x`

Note this solution has `4` constants of integration and `4` linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution