What is the solution for y^{''}-8y^'+25y=0,y(0)=e^{4x}cos(3x) ?

The solution for y^{''}-8y^'+25y=0,y(0)=e^{4x}cos(3x) is y=e^{4x}(c_{1}cos(3x)+c_{2}sin(3x))

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y^{''}-8y^'+25y=0,y(0)=e^{4x}cos(3x)

y^{^{''}} - 8 y^{^{'}} + 25 y = 0, y (0) = e^{4 x} cos (3 x)

y = e^{4 x} (c_{1} cos (3 x) + c_{2} sin (3 x))

Solution steps

y^{''} (x) - 8 y^{'} (x) + 25 y = 0

Solve linear ODE:

y = e^{4 x} (c_{1} cos (3 x) + c_{2} sin (3 x))

y = e^{4 x} (c_{1} cos (3 x) + c_{2} sin (3 x))

What is the solution for y^{''}-8y^'+25y=0,y(0)=e^{4x}cos(3x) ?
The solution for y^{''}-8y^'+25y=0,y(0)=e^{4x}cos(3x) is y=e^{4x}(c_{1}cos(3x)+c_{2}sin(3x))