What is the solution for y^{''}-y^'+20y=0 ?

The solution for y^{''}-y^'+20y=0 is y=e^{t/2}(c_{1}cos((sqrt(79)t)/2)+c_{2}sin((sqrt(79)t)/2))

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y^{''}-y^'+20y=0

y^{^{''}} - y^{^{'}} + 20 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{79 t}{2}) + c_{2} sin (\frac{79 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{79 t}{2}) + c_{2} sin (\frac{79 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{79 t}{2}) + c_{2} sin (\frac{79 t}{2}))

x \to 3 + lim (\frac{5}{x - 3})

(x - 2)^{6}

x \to 1 lim (3 x - 2)

\int \frac{( sin ( x ) ) ^{3}}{1 + ( cos ( x ) ) ^{3}} d x

\frac{d}{d x} (\frac{5}{5 - x})

What is the solution for y^{''}-y^'+20y=0 ?
The solution for y^{''}-y^'+20y=0 is y=e^{t/2}(c_{1}cos((sqrt(79)t)/2)+c_{2}sin((sqrt(79)t)/2))