What is the solution for y^{''}+y=6e^{-8x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=6e^{-8x},y(0)=0,y^'(0)=0 is y=-6/65 cos(x)+48/65 sin(x)+6/65 e^{-8x}

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y^{''}+y=6e^{-8x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 6 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{6}{65} cos (x) + \frac{48}{65} sin (x) + \frac{6}{65} e^{- 8 x}

Solution steps

y^{''} (x) + y = 6 e^{- 8 x}

Solve linear ODE:

y = - \frac{6}{65} cos (x) + \frac{48}{65} sin (x) + \frac{6}{65} e^{- 8 x}

y = - \frac{6}{65} cos (x) + \frac{48}{65} sin (x) + \frac{6}{65} e^{- 8 x}

y^{^{''}} + 9 y = x^{2} e^{3 x}

6 y^{^{''}} - y^{^{'}} - 7 y = - 7

y^{^{''}} - y^{^{'}} - 6 y = 58 cos (3 x)

y^{^{''}} - 9 y = (x^{2} - 7) sin (3 x)

y^{^{''}} + 36 y = csc^{2} (6 x)

What is the solution for y^{''}+y=6e^{-8x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=6e^{-8x},y(0)=0,y^'(0)=0 is y=-6/65 cos(x)+48/65 sin(x)+6/65 e^{-8x}