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

n = 1 \sum \infty ln (\frac{1}{n ^{2}})

\frac{d}{d x} (2 x (3 x^{2} - 7)^{3})

x \to 3 lim (\frac{x ^{2} - 1}{x - 3})

\int \frac{z}{z - 1} d z

\int \frac{2}{x ^{2} - x} d x

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