What is the solution for 4y^{''}-4y^'+26y=0 ?

The solution for 4y^{''}-4y^'+26y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))

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4y^{''}-4y^'+26y=0

4 y^{^{''}} - 4 y^{^{'}} + 26 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

ln (x + 1)

\int \frac{4 x}{2 + 2 x ^{2}} d x

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What is the solution for 4y^{''}-4y^'+26y=0 ?
The solution for 4y^{''}-4y^'+26y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))