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

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

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

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

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 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

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = - 2

12 y^{^{''}} - 5 y^{^{'}} - 2 y = 0

y^{^{''}} - 8 y^{^{'}} + 17 y = 0

x^{^{''}} + 4 x = 0

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