What is the solution for 25y^{''}-10y^'+y=0 ?

The solution for 25y^{''}-10y^'+y=0 is y=c_{1}e^{t/5}+c_{2}e^{t/5}t

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25y^{''}-10y^'+y=0

25 y^{^{''}} - 10 y^{^{'}} + y = 0

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{5}} t

Solution steps

25 y^{''} (t) - 10 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{5}} t

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{5}} t

lo g_{3} (2 x + 5)

\int \frac{sec ^{2} ( t )}{1 + tan ( t )} d t

\int (\frac{2 y ^{\frac{1}{2}}}{y ^{3}}) d^{0} x

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4})

(1 + ln (x) + \frac{y}{x}) d x + (- 1 + ln (x)) d y = 0

What is the solution for 25y^{''}-10y^'+y=0 ?
The solution for 25y^{''}-10y^'+y=0 is y=c_{1}e^{t/5}+c_{2}e^{t/5}t