What is the solution for y^{''}+y=3e^{-8x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=3e^{-8x},y(0)=0,y^'(0)=0 is y=-3/65 cos(x)+24/65 sin(x)+3/65 e^{-8x}

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y^{''}+y=3e^{-8x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 3 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{3}{65} cos (x) + \frac{24}{65} sin (x) + \frac{3}{65} e^{- 8 x}

Solution steps

y^{''} (x) + y = 3 e^{- 8 x}

Solve linear ODE:

y = - \frac{3}{65} cos (x) + \frac{24}{65} sin (x) + \frac{3}{65} e^{- 8 x}

y = - \frac{3}{65} cos (x) + \frac{24}{65} sin (x) + \frac{3}{65} e^{- 8 x}

x \to 4 lim (16 - x^{2})

x \to - 2 lim (5)

y = x^{3} - 4 x + 3

\int 4 x + 5 d x

x \to 1 lim (\frac{tan ( x )}{x})

What is the solution for y^{''}+y=3e^{-8x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=3e^{-8x},y(0)=0,y^'(0)=0 is y=-3/65 cos(x)+24/65 sin(x)+3/65 e^{-8x}