What is the solution for y^{''}-y^'+1/4 y=0 ?

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

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y^{''}-y^'+1/4 y=0

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 0

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

Solution steps

y^{''} (t) - y^{'} (t) + \frac{1}{4} y = 0

Solve linear ODE:

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

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

x \to - 1 + lim (\frac{3 x - 5}{x + 1})

- \frac{3}{x}

(arctan (cos (x)))^{^{'}}

y^{^{''}} - y^{^{'}} - 6 y = 20 e^{- 2 x}

\frac{\partial}{\partial x} (8 x y^{5} + x^{8} y^{2})

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