What is the solution for 9y^{''}+4y=3x+12e-x,y(0)=1,y^'(0)=0 ?

The solution for 9y^{''}+4y=3x+12e-x,y(0)=1,y^'(0)=0 is y=(1-3e)cos((2x)/3)-3/4 sin((2x)/3)+x/2+3e

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9y^{''}+4y=3x+12e-x,y(0)=1,y^'(0)=0

9 y^{^{''}} + 4 y = 3 x + 12 e - x, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - 3 e) cos (\frac{2 x}{3}) - \frac{3}{4} sin (\frac{2 x}{3}) + \frac{x}{2} + 3 e

Solution steps

9 y^{''} (x) + 4 y = 3 x + 12 e - x

Solve linear ODE:

y = (1 - 3 e) cos (\frac{2 x}{3}) - \frac{3}{4} sin (\frac{2 x}{3}) + \frac{x}{2} + 3 e

y = (1 - 3 e) cos (\frac{2 x}{3}) - \frac{3}{4} sin (\frac{2 x}{3}) + \frac{x}{2} + 3 e

y^{^{''}} - y = 4 cos (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 3 y^{^{'}} + 2 y = a, y (0) = 2, y^{^{'}} (0) = - 6

d (x) = \frac{2500}{30 ( 0.3 + 0.03 )}

y^{^{''}} = 2 y - x

y^{^{''}} - y \cdot a + g = 0

What is the solution for 9y^{''}+4y=3x+12e-x,y(0)=1,y^'(0)=0 ?
The solution for 9y^{''}+4y=3x+12e-x,y(0)=1,y^'(0)=0 is y=(1-3e)cos((2x)/3)-3/4 sin((2x)/3)+x/2+3e