What is the solution for y^{''}+y=7e^{-5x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=7e^{-5x},y(0)=0,y^'(0)=0 is y=-7/26 cos(x)+35/26 sin(x)+7/26 e^{-5x}

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y^{''}+y=7e^{-5x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 7 e^{- 5 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{7}{26} cos (x) + \frac{35}{26} sin (x) + \frac{7}{26} e^{- 5 x}

Solution steps

y^{''} (x) + y = 7 e^{- 5 x}

Solve linear ODE:

y = - \frac{7}{26} cos (x) + \frac{35}{26} sin (x) + \frac{7}{26} e^{- 5 x}

y = - \frac{7}{26} cos (x) + \frac{35}{26} sin (x) + \frac{7}{26} e^{- 5 x}

\frac{\partial}{\partial x} (4 y x^{3} - 15 y^{2} x^{4})

\frac{d}{d t} (8 sin (t))

y = \frac{- 4 x}{x ^{2} + 1}

5 x^{3} - 6 x^{2} + 12 x - 3, 3

\frac{d}{d x} (4 x^{3} + 3 x^{2} - 6 x + 1)

What is the solution for y^{''}+y=7e^{-5x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=7e^{-5x},y(0)=0,y^'(0)=0 is y=-7/26 cos(x)+35/26 sin(x)+7/26 e^{-5x}