What is the solution for y^{''}+y=8e^{-2x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=8e^{-2x},y(0)=0,y^'(0)=0 is y=-8/5 cos(x)+16/5 sin(x)+8/5 e^{-2x}

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y^{''}+y=8e^{-2x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 8 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{8}{5} cos (x) + \frac{16}{5} sin (x) + \frac{8}{5} e^{- 2 x}

Solution steps

y^{''} (x) + y = 8 e^{- 2 x}

Solve linear ODE:

y = - \frac{8}{5} cos (x) + \frac{16}{5} sin (x) + \frac{8}{5} e^{- 2 x}

y = - \frac{8}{5} cos (x) + \frac{16}{5} sin (x) + \frac{8}{5} e^{- 2 x}

f (x) = 0.27 x^{2} - 8.6 x + 93

\frac{d}{d x} (\frac{6 x ^{2}}{2 y + cos ( y )})

\frac{d}{d y} (y^{3})

\int x a^{2 x} d x

\frac{\partial}{\partial x} (cos (x) sec^{2} (y))

What is the solution for y^{''}+y=8e^{-2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=8e^{-2x},y(0)=0,y^'(0)=0 is y=-8/5 cos(x)+16/5 sin(x)+8/5 e^{-2x}