What is the solution for y^{''}-2y^'+10y=0,y(0)=4,y^'(0)=1 ?

The solution for y^{''}-2y^'+10y=0,y(0)=4,y^'(0)=1 is y=e^t(4cos(3t)-sin(3t))

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y^{''}-2y^'+10y=0,y(0)=4,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y = e^{t} (4 cos (3 t) - sin (3 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{t} (4 cos (3 t) - sin (3 t))

y = e^{t} (4 cos (3 t) - sin (3 t))

x \to - 9 - lim (x^{2} + \frac{3}{x + 9})

\frac{\partial}{\partial y} (\frac{- y}{2 x ^{2} + y ^{2}})

\int (3 z + z^{2}) sin (z) d z

\frac{x - x}{x ^{\frac{1}{5}}}

\int \frac{in ( x )}{x ^{2}} d x

What is the solution for y^{''}-2y^'+10y=0,y(0)=4,y^'(0)=1 ?
The solution for y^{''}-2y^'+10y=0,y(0)=4,y^'(0)=1 is y=e^t(4cos(3t)-sin(3t))