What is the solution for 9y^{''}-6y^'+y=0 ?

The solution for 9y^{''}-6y^'+y=0 is y=c_{1}e^{t/3}+c_{2}e^{t/3}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9y^{''}-6y^'+y=0

9 y^{^{''}} - 6 y^{^{'}} + y = 0

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t

Solution steps

9 y^{''} (t) - 6 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t

n = 1 \sum \infty (\frac{1}{7})^{n}

4 x - x^{2}, x^{2}, 2, 0

x \to - 6 - lim (\frac{1}{( x + 6 ) ^{6}})

f (x) = e^{x} + x

\frac{d y}{d x} = \frac{1 + cos ( 4 x )}{y e ^{y^{2}}}

What is the solution for 9y^{''}-6y^'+y=0 ?
The solution for 9y^{''}-6y^'+y=0 is y=c_{1}e^{t/3}+c_{2}e^{t/3}t