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

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

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

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

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

Solution steps

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

y^{^{'''}} + y^{^{''}} + 2 y^{^{'}} + 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + 4 f (x) = 0

y^{^{''}} = - 16 y^{^{'}}

7 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

4 y^{^{''}} - 72 y^{^{'}} + 324 y = 0

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