What is the solution for y^{''}+4y+5y=35e^{-4x},y(0)=-6,y^'(0)=1 ?

The solution for y^{''}+4y+5y=35e^{-4x},y(0)=-6,y^'(0)=1 is y=-37/5 cos(3x)+11/5 sin(3x)+7/5 e^{-4x}

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y^{''}+4y+5y=35e^{-4x},y(0)=-6,y^'(0)=1

y^{^{''}} + 4 y + 5 y = 35 e^{- 4 x}, y (0) = - 6, y^{^{'}} (0) = 1

y = - \frac{37}{5} cos (3 x) + \frac{11}{5} sin (3 x) + \frac{7}{5} e^{- 4 x}

Solution steps

y^{''} (x) + 4 y + 5 y = 35 e^{- 4 x}

Solve linear ODE:

y = - \frac{37}{5} cos (3 x) + \frac{11}{5} sin (3 x) + \frac{7}{5} e^{- 4 x}

y = - \frac{37}{5} cos (3 x) + \frac{11}{5} sin (3 x) + \frac{7}{5} e^{- 4 x}

What is the solution for y^{''}+4y+5y=35e^{-4x},y(0)=-6,y^'(0)=1 ?
The solution for y^{''}+4y+5y=35e^{-4x},y(0)=-6,y^'(0)=1 is y=-37/5 cos(3x)+11/5 sin(3x)+7/5 e^{-4x}