# ((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

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### 1 Answer

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.`