What is the solution for 2y^{''}-9y^'+4y=0 ?

The solution for 2y^{''}-9y^'+4y=0 is y=c_{1}e^{4t}+c_{2}e^{t/2}

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2y^{''}-9y^'+4y=0

2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

y = c_{1} e^{4 t} + c_{2} e^{\frac{t}{2}}

Solution steps

2 y^{''} (t) - 9 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{\frac{t}{2}}

y = c_{1} e^{4 t} + c_{2} e^{\frac{t}{2}}

f (x) = x^{6} e^{x}

y = e^{- 2 x} \cdot tan (8 x)

x \to 0 lim (\frac{3 sin ( 3 x )}{x})

x, sin (x), [\frac{π}{2}, π]

y^{^{''}} + 3 y = 1

What is the solution for 2y^{''}-9y^'+4y=0 ?
The solution for 2y^{''}-9y^'+4y=0 is y=c_{1}e^{4t}+c_{2}e^{t/2}