What is the solution for y^{''}-10y^'+34y=0,y(0)=-2,y^'(0)=-3 ?

The solution for y^{''}-10y^'+34y=0,y(0)=-2,y^'(0)=-3 is y=e^{5t}(-2cos(3t)+7/3 sin(3t))

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y^{''}-10y^'+34y=0,y(0)=-2,y^'(0)=-3

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (0) = - 2, y^{^{'}} (0) = - 3

y = e^{5 t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

Solution steps

y^{''} (t) - 10 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{5 t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

y = e^{5 t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

\frac{3}{x}, \frac{3}{x ^{2}}, 6

y = 5, y = x^{2} + 1

(x - 2)^{2} - 3 x

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

What is the solution for y^{''}-10y^'+34y=0,y(0)=-2,y^'(0)=-3 ?
The solution for y^{''}-10y^'+34y=0,y(0)=-2,y^'(0)=-3 is y=e^{5t}(-2cos(3t)+7/3 sin(3t))