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

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

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

y^{^{''}} - 8 y^{^{'}} = 20 y, y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{8} x^{^{''}} + \frac{5}{4} x^{^{'}} + \frac{1}{3} x = 0

y^{^{'''}} - 3 y^{^{''}} - 9 y^{^{'}} - 5 y = 0

θ^{^{''}} (t) + \frac{a}{b} \cdot θ (t) = 0

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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