What is the solution for y^{''}-2y^'+50y=0,y(0)= 3/2 ,y^'(0)=-9 ?

The solution for y^{''}-2y^'+50y=0,y(0)= 3/2 ,y^'(0)=-9 is y=e^t(3/2 cos(7t)-3/2 sin(7t))

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y^{''}-2y^'+50y=0,y(0)= 3/2 ,y^'(0)=-9

y^{^{''}} - 2 y^{^{'}} + 50 y = 0, y (0) = \frac{3}{2}, y^{^{'}} (0) = - 9

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 50 y = 0

Solve linear ODE:

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

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

\int_{0}^{1 - x} 24 y (1 - x - y) d y

4 y^{2}

\int x \cdot sin^{3} (x) d x

y = \frac{6 x}{x + 3}, (- 1, - 3)

f (x) = \frac{2 x}{x ^{2} - 8}

What is the solution for y^{''}-2y^'+50y=0,y(0)= 3/2 ,y^'(0)=-9 ?
The solution for y^{''}-2y^'+50y=0,y(0)= 3/2 ,y^'(0)=-9 is y=e^t(3/2 cos(7t)-3/2 sin(7t))