What is the solution for y^{''}-y^'+0.25y=0 ?

The solution for y^{''}-y^'+0.25y=0 is y=c_{1}e^{t/2}+c_{2}e^{t/2}t

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y^{''}-y^'+0.25y=0

y^{^{''}} - y^{^{'}} + 0.25 y = 0

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

Solution steps

y^{''} (t) - y^{'} (t) + 0.25 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

\int e^{4 x} sin (b x) d x

f (x) = x + 9

\int 5 x^{\frac{3}{2}} d x

x \to \infty lim ((x) (e^{\frac{1}{x}}))

t \to 1 lim (t + 4)

What is the solution for y^{''}-y^'+0.25y=0 ?
The solution for y^{''}-y^'+0.25y=0 is y=c_{1}e^{t/2}+c_{2}e^{t/2}t