What is the solution for y^{''}-10y^'+50y=0,y(0)=5,y^'(0)=-2 ?

The solution for y^{''}-10y^'+50y=0,y(0)=5,y^'(0)=-2 is y=e^{5t}(5cos(5t)-27/5 sin(5t))

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y^{''}-10y^'+50y=0,y(0)=5,y^'(0)=-2

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 5, y^{^{'}} (0) = - 2

y = e^{5 t} (5 cos (5 t) - \frac{27}{5} sin (5 t))

Solution steps

y^{''} (t) - 10 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{5 t} (5 cos (5 t) - \frac{27}{5} sin (5 t))

y = e^{5 t} (5 cos (5 t) - \frac{27}{5} sin (5 t))

5 y^{^{''}} + 30 y = 0, y (0) = 5, y^{^{'}} (0) = 5

\int \frac{1}{a ^{2} + u ^{2}} d u

\frac{\partial}{\partial y} ((2 x - 5 y)^{2})

(10 y^{2} - x y) d x + (x^{2}) d y = 0

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

What is the solution for y^{''}-10y^'+50y=0,y(0)=5,y^'(0)=-2 ?
The solution for y^{''}-10y^'+50y=0,y(0)=5,y^'(0)=-2 is y=e^{5t}(5cos(5t)-27/5 sin(5t))