What is the solution for 12y^{''}+5y^'-3y=e^x ?

The solution for 12y^{''}+5y^'-3y=e^x is y=c_{1}e^{x/3}+c_{2}e^{-(3x)/4}+1/14 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12y^{''}+5y^'-3y=e^x

12 y^{^{''}} + 5 y^{^{'}} - 3 y = e^{x}

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{- \frac{3 x}{4}} + \frac{1}{14} e^{x}

Solution steps

12 y^{''} (x) + 5 y^{'} (x) - 3 y = e^{x}

Solve linear ODE:

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{- \frac{3 x}{4}} + \frac{1}{14} e^{x}

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{- \frac{3 x}{4}} + \frac{1}{14} e^{x}

y^{^{''}} - 3 y^{^{'}} - 2 y = 9 e^{- x}

(D^{3} + D^{2} + 5 D + 5) y = 5 cos (2 x)

y^{^{''}} + 4 y = x (sin (2 x))

y^{^{''}} - 2 y^{^{'}} - 3 y = 32 e^{- x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 sin (4 t)

What is the solution for 12y^{''}+5y^'-3y=e^x ?
The solution for 12y^{''}+5y^'-3y=e^x is y=c_{1}e^{x/3}+c_{2}e^{-(3x)/4}+1/14 e^x