What is the solution for y^{''}-12y^'+37y=0,y(0)=6,y^'(0)=8 ?

The solution for y^{''}-12y^'+37y=0,y(0)=6,y^'(0)=8 is y=e^{6t}(6cos(t)-28sin(t))

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y^{''}-12y^'+37y=0,y(0)=6,y^'(0)=8

y^{^{''}} - 12 y^{^{'}} + 37 y = 0, y (0) = 6, y^{^{'}} (0) = 8

y = e^{6 t} (6 cos (t) - 28 sin (t))

Solution steps

y^{''} (t) - 12 y^{'} (t) + 37 y = 0

Solve linear ODE:

y = e^{6 t} (6 cos (t) - 28 sin (t))

y = e^{6 t} (6 cos (t) - 28 sin (t))

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\int \frac{x ^{2} + 12 x + 12}{x ^{3} - 4 x} d x

\int x^{2} csc^{2} (x) cot (x) d x

\int + sin^{3} (x) cos^{14} (x) d x

\frac{3 x}{x - \frac{3}{x}}

What is the solution for y^{''}-12y^'+37y=0,y(0)=6,y^'(0)=8 ?
The solution for y^{''}-12y^'+37y=0,y(0)=6,y^'(0)=8 is y=e^{6t}(6cos(t)-28sin(t))