What is the solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-3,y^'(0)=1 ?

The solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-3,y^'(0)=1 is y=e^{-2x}(-10cos(x)+9sin(x))+7e^{-4x}

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y^{''}+4y^'+5y=35e^{-4x},y(0)=-3,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4 x}, y (0) = - 3, y^{^{'}} (0) = 1

y = e^{- 2 x} (- 10 cos (x) + 9 sin (x)) + 7 e^{- 4 x}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (- 10 cos (x) + 9 sin (x)) + 7 e^{- 4 x}

y = e^{- 2 x} (- 10 cos (x) + 9 sin (x)) + 7 e^{- 4 x}

4 x^{^{'}} + x = 5

\frac{d y}{d t} + 0.1 t y = 2 t, y (0) = 3

y^{\frac{1}{2}} y^{^{'}} + y^{\frac{3}{2}} = 1, y (0) = 4

\frac{\partial}{\partial x} (x \cdot y \cdot z)

y = x^{4} + 7 e^{x}, (0, 7)

What is the solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-3,y^'(0)=1 ?
The solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-3,y^'(0)=1 is y=e^{-2x}(-10cos(x)+9sin(x))+7e^{-4x}