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))

(- 3, - 1), (- 2, - 2)

\int \frac{1}{5 x} d x

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

3 y^{^{''}} + 13 y^{^{'}} + 4 y = 0, y (0) = - 2, y^{^{'}} (0) = - 2

\int (5 x^{3} - 3 x^{2} + 7) d 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))