What is the solution for 9y^{''}-12y^'+4y=0,y(0)=e^{2/3 x} ?

The solution for 9y^{''}-12y^'+4y=0,y(0)=e^{2/3 x} is y=c_{1}e^{(2x)/3}+c_{2}e^{(2x)/3}x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9y^{''}-12y^'+4y=0,y(0)=e^{2/3 x}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2}{3} x}

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

Solution steps

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

Solve linear ODE:

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

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

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

2 y - 3 y^{^{''}} = 0

y^{^{''}} + 4 y^{^{'}} - 7 y = 0

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

x^{^{''}} + 2 x^{^{'}} + 6 x = 0

What is the solution for 9y^{''}-12y^'+4y=0,y(0)=e^{2/3 x} ?
The solution for 9y^{''}-12y^'+4y=0,y(0)=e^{2/3 x} is y=c_{1}e^{(2x)/3}+c_{2}e^{(2x)/3}x