What is the solution for (d-1)^4y=e^x ?

The solution for (d-1)^4y=e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+c_{4}x^3e^x+(e^xx^4)/(24)

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(d-1)^4y=e^x

(d - 1)^{4} y = e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + c_{4} x^{3} e^{x} + \frac{e ^{x} x ^{4}}{24}

Solution steps

(d - 1)^{4} y = e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + c_{4} x^{3} e^{x} + \frac{e ^{x} x ^{4}}{24}

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + c_{4} x^{3} e^{x} + \frac{e ^{x} x ^{4}}{24}

y^{^{''}} + 4 y = 2 e^{x}

(d^{3} + 4 d^{2} + 4 d) \cdot y = x^{2} \cdot e^{- 2 x}

(d - 2)^{2} y = 8 (e^{2} x + sin^{2} (x) + x^{2})

[d^{2} - 4 d + 5] y = e^{3} x

(d - 3)^{2} \cdot y = 2 e^{4 x}

What is the solution for (d-1)^4y=e^x ?
The solution for (d-1)^4y=e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+c_{4}x^3e^x+(e^xx^4)/(24)