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

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

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

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^{^{''}} + 2 y^{^{'}} + (1 + \frac{9}{4} π^{2}) y = 0

y^{^{''}} + 16 y = 0, y (0) = cos (4 x)

y^{^{''}} - y^{^{'}} - y = 0, y^{1} (0) = e^{x}, y^{2} (0) = x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

y^{^{''}} + \frac{49}{\frac{4}{32}} y = 0

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