What is the solution for y^{''}-10y^'+50y=0,y(0)=3,y^'(0)=-5 ?

The solution for y^{''}-10y^'+50y=0,y(0)=3,y^'(0)=-5 is y=e^{5t}(3cos(5t)-4sin(5t))

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y^{''}-10y^'+50y=0,y(0)=3,y^'(0)=-5

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

y = e^{5 t} (3 cos (5 t) - 4 sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{5 t} (3 cos (5 t) - 4 sin (5 t))

y = e^{5 t} (3 cos (5 t) - 4 sin (5 t))

\frac{d ^{2} f}{d x ^{2}} + f (x) = 0

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = - \frac{1}{6}

y^{^{''}} - 2 \cdot y^{^{'}} + y = 0

25 y^{^{''}} + 40 y^{^{'}} + 16 y = 0

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = 2, y^{^{'}} (0) = 4

What is the solution for y^{''}-10y^'+50y=0,y(0)=3,y^'(0)=-5 ?
The solution for y^{''}-10y^'+50y=0,y(0)=3,y^'(0)=-5 is y=e^{5t}(3cos(5t)-4sin(5t))