What is the solution for y^{''}-8y^'+17y=0,y(0)=4,y^'(0)=-1 ?

The solution for y^{''}-8y^'+17y=0,y(0)=4,y^'(0)=-1 is y=e^{4t}(4cos(t)-17sin(t))

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y^{''}-8y^'+17y=0,y(0)=4,y^'(0)=-1

y^{^{''}} - 8 y^{^{'}} + 17 y = 0, y (0) = 4, y^{^{'}} (0) = - 1

y = e^{4 t} (4 cos (t) - 17 sin (t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{4 t} (4 cos (t) - 17 sin (t))

y = e^{4 t} (4 cos (t) - 17 sin (t))

\int \frac{t}{t ^{4} + 3} d t

x \to 1 - lim (arccos (- x^{3}))

(\frac{x ^{2} + 2}{x ^{2} - 2})^{5}

\frac{d}{d x} (\frac{1}{3 x ^{2} - 1})

20 x - 16 x^{2}

What is the solution for y^{''}-8y^'+17y=0,y(0)=4,y^'(0)=-1 ?
The solution for y^{''}-8y^'+17y=0,y(0)=4,y^'(0)=-1 is y=e^{4t}(4cos(t)-17sin(t))