What is the solution for 2y^{''}-3y^'+4y=0 ?

The solution for 2y^{''}-3y^'+4y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))

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2y^{''}-3y^'+4y=0

2 y^{^{''}} - 3 y^{^{'}} + 4 y = 0

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

\frac{\partial}{\partial z} (- x^{3} yz e^{z})

\frac{d}{d x} (1000 e^{- \frac{x}{10}})

\int x^{2} (8 x^{3} + 7)^{2} d x

\frac{d}{d x} (arctan (7^{x}))

\int \frac{( e ^{a r c t a n (x)} )}{1 + x ^{2}} d x

What is the solution for 2y^{''}-3y^'+4y=0 ?
The solution for 2y^{''}-3y^'+4y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))