((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= x+sinx+e^x`` Linear non-homogeneous. solve for the particular solution

You need to evaluate the particular solution to the given linear non homogeneous differential equation, hence, you need to search the solutions to following equations, such that:

`{(((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= x),(((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= sin x),(((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= e^x):}` For the equation `((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= x` , you need to try the solution `y_(p1) = Ax + B` , such that:

`(dy)/(dx) = A`

`(d^2y)/(dx^2) = (d^3y)/(dx^3) = (d^4y)/(dx^4) = (d^5y)/(dx^5) = 0`

`80A + 32(Ax + B) = x`

`32Ax + 80A + 32B = x`

Equating the coefficients of like powers yields:

`32A = 1 => A = 1/32`

`80A + 32B = 0 => B = -80A/32 => B = -80/(32^2)=>B = -10/(4*32) => B = -5/(64)`

Hence, evaluating `y_(p1)` yields `y_(p1) = 1/32x - 5/64` .

For the equation `((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= sin x` , you need to try the solution `y_(p2) = C cos x + D sin x` , such that:

`(dy)/(dx) = -C sin x + D cos x`

`(d^2y)/(dx^2) = -C cos x - D sin x`

`(d^3y)/(dx^3) = C sin x - D cos x`

`(d^4y)/(dx^4) = C cos x + D sin x =y_(p2)`

`(d^5y)/(dx^5) = (dy)/(dx) = -C sin x + D cos x`

`-C sin x + D cos x + 10 C cos x + 10D sin x + 40C sin x - 40D cos x - 80C cos x - 80D sin x- 80C sin x + 80D cos x + 32C cos x + 32D sin x = sin x`

`sin x(-C + 10D + 40C - 80D - 80C + 32D) + cos x(D + 10C - 40D - 80C + 80D + 32C) = sin x`

Equating the coefficients of like parts yields:

`-C + 10D + 40C - 80D - 80C + 32D = 1 => -41C - 38D = 1`

`-38C + 41D = 0 => D = 38C/41`

Multiplying the second equation by 38 and adding the resulted equation to the first one, multiplied by 41, you may reduce the variable D, such that:

`-41*41C - 38*41D - 38*38C + 41*38D = 41`

`1681C + 1444C = -41`

`3125C = -41 => C = -41/3125 => D = -38/3125`

Hence, evaluating `y_(p2)` yields `y_(p2) = (-41/3125) cos x + (-38/3125) sin x.`

For the equation `((d^5y)/(dx^5))+(10(d^4y)/(dx^4))+(40(d^3y)/(dx^3))+(80(d^2y)/(dx^2))+(80(dy)/(dx))+32y= e^x` , you need to try the solution `y_(p3) = E*e^x` , such that:

`(dy)/(dx) = (d^2y)/(dx^2) = (d^3y)/(dx^3) = (d^4y)/(dx^4) = (d^5y)/(dx^5) = E*e^x`

`E*e^x+10E*e^x+40E*e^x+80E*e^x+80E*e^x+32E*e^x= e^x`

`243E*e^x = e^x => E = 1/243`

Hence, evaluating `y_(p3)` yields `y_(p3) = (1/243)e^x.`

Hence, using the principle of superposition, yields that the particular solution to the given equation is `y_p = y_(p1) + y_(p2) + y_(p3) = 1/32x - 5/64 + (-41/3125) cos x + (-38/3125) sin x + (1/243)e^x.`

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