What is the solution for y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x) ?

The solution for y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x) is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))

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y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y^{1} (0) = e^{x} cos (2 x)

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x))

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = 0

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x))

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x))

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0, y (π) = 1

y^{^{'''}} - 3 y^{^{''}} - 5 y^{^{'}} + 39 y = 0

y^{^{''}} + 29.4 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{y ^{^{''}} + y + y}{x} = 0

What is the solution for y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x) ?
The solution for y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x) is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))