What is the answer to solvefor y,(d^3+4d)y=2t+t^2 ?

The answer to solvefor y,(d^3+4d)y=2t+t^2 is y=(2t+t^2)/(d^3+4d);\quad d\ne 0,d\ne 2i,d\ne-2i

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solvefor y,(d^3+4d)y=2t+t^2

solve for

y, (d^{3} + 4 d) y = 2 t + t^{2}

y = \frac{2 t + t ^{2}}{d ^{3} + 4 d}; d \neq = 0, d \neq = 2 i, d \neq = - 2 i

Solution steps

(d^{3} + 4 d) y = 2 t + t^{2}

Divide both sides by

d^{3} + 4 d; d \neq = 0, d \neq = 2 i, d \neq = - 2 i

y = \frac{2 t + t ^{2}}{d ^{3} + 4 d}; d \neq = 0, d \neq = 2 i, d \neq = - 2 i

y^{^{''}} - 9 y = x

y^{^{''}} + y = 2 \cdot e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 10 x e^{3 x} + 17 e^{3 x}

y^{^{''}} - 4 y = x + sin (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} sin (x)

What is the answer to solvefor y,(d^3+4d)y=2t+t^2 ?
The answer to solvefor y,(d^3+4d)y=2t+t^2 is y=(2t+t^2)/(d^3+4d);\quad d\ne 0,d\ne 2i,d\ne-2i