What is the solution for (d^4+4d^2)y=x^2+1 ?

The solution for (d^4+4d^2)y=x^2+1 is y=c_{1}+c_{2}x+c_{3}cos(2x)+c_{4}sin(2x)+(x^4)/(48)+(x^2)/(16)

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(d^4+4d^2)y=x^2+1

(d^{4} + 4 d^{2}) y = x^{2} + 1

y = c_{1} + c_{2} x + c_{3} cos (2 x) + c_{4} sin (2 x) + \frac{x ^{4}}{48} + \frac{x ^{2}}{16}

Solution steps

(d^{4} + 4 d^{2}) y = x^{2} + 1

Solve linear ODE:

y = c_{1} + c_{2} x + c_{3} cos (2 x) + c_{4} sin (2 x) + \frac{x ^{4}}{48} + \frac{x ^{2}}{16}

y = c_{1} + c_{2} x + c_{3} cos (2 x) + c_{4} sin (2 x) + \frac{x ^{4}}{48} + \frac{x ^{2}}{16}

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} (x^{2} - 4 x + 5)

y^{^{''}} + 2 y = e^{t} + 2

y^{^{''}} - 5 y^{^{'}} + 9 y = x e^{x}

y^{^{''}} - y^{^{'}} - 30 y = 2 x + 4

What is the solution for (d^4+4d^2)y=x^2+1 ?
The solution for (d^4+4d^2)y=x^2+1 is y=c_{1}+c_{2}x+c_{3}cos(2x)+c_{4}sin(2x)+(x^4)/(48)+(x^2)/(16)