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

The solution for y^{''}-y^'+2y=0,y(0)=5 is y=e^{t/2}(5cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

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

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

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

x^{^{''}} + c^{2} \cdot x = 0

y^{^{'''}} - 8 y^{^{''}} - y^{^{'}} + 8 y = 0

a \frac{d ^{2} y}{d x ^{2}} + y b^{2} = 0

y^{^{''}} + \frac{48}{2} y = 0

y^{^{''}} - y^{^{'}} + 0.25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

What is the solution for y^{''}-y^'+2y=0,y(0)=5 ?
The solution for y^{''}-y^'+2y=0,y(0)=5 is y=e^{t/2}(5cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))