What is the solution for y^{''}-30y^'+225y=0 ?

The solution for y^{''}-30y^'+225y=0 is y=c_{1}e^{15t}+c_{2}te^{15t}

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y^{''}-30y^'+225y=0

y^{^{''}} - 30 y^{^{'}} + 225 y = 0

y = c_{1} e^{15 t} + c_{2} t e^{15 t}

Solution steps

y^{''} (t) - 30 y^{'} (t) + 225 y = 0

Solve linear ODE:

y = c_{1} e^{15 t} + c_{2} t e^{15 t}

y = c_{1} e^{15 t} + c_{2} t e^{15 t}

y \to \frac{1}{2} lim ((16 y^{2} - 4)^{4})

x \to o lim (x^{x})

\frac{d}{d y} (2 x y^{4} e^{y})

f (x) 6 x^{4} - 10 x^{3} - 5 x^{2} + 16 x - 45

f (x) = \frac{2}{3 x + 1}, at x = - 1

What is the solution for y^{''}-30y^'+225y=0 ?
The solution for y^{''}-30y^'+225y=0 is y=c_{1}e^{15t}+c_{2}te^{15t}