What is the solution for y^{''}-8y^'+18y=0,y(0)=1,y^'(0)=-4 ?

The solution for y^{''}-8y^'+18y=0,y(0)=1,y^'(0)=-4 is y=e^{4t}(cos(sqrt(2)t)-4sqrt(2)sin(sqrt(2)t))

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y^{''}-8y^'+18y=0,y(0)=1,y^'(0)=-4

y^{^{''}} - 8 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

y = e^{4 t} (cos (2 t) - 42 sin (2 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = e^{4 t} (cos (2 t) - 42 sin (2 t))

y = e^{4 t} (cos (2 t) - 42 sin (2 t))

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\frac{d y}{d x} = - \frac{1}{3} x^{3}

\frac{d ^{2}}{d x ^{2}} (x + \frac{1}{x})

\int \frac{cos ( 2 x )}{cos ( x )} d x

What is the solution for y^{''}-8y^'+18y=0,y(0)=1,y^'(0)=-4 ?
The solution for y^{''}-8y^'+18y=0,y(0)=1,y^'(0)=-4 is y=e^{4t}(cos(sqrt(2)t)-4sqrt(2)sin(sqrt(2)t))