What is the solution for 9y^{''}-12y^'+4y=0 ?

The solution for 9y^{''}-12y^'+4y=0 is y=c_{1}e^{(2t)/3}+c_{2}e^{(2t)/3}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9y^{''}-12y^'+4y=0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0

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

Solution steps

9 y^{''} (t) - 12 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

\frac{5 x ^{2}}{( 3 - 4 x ) ^{4}}

sin (3 t) cos (3 t)

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

2 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

\frac{d y}{d x} = \frac{y}{x ^{3} + 3 x ^{2} + 2 x}

What is the solution for 9y^{''}-12y^'+4y=0 ?
The solution for 9y^{''}-12y^'+4y=0 is y=c_{1}e^{(2t)/3}+c_{2}e^{(2t)/3}t