What is the solution for y^{''}-y=2ln(x) ?

The solution for y^{''}-y=2ln(x) is y=c_{1}e^x+c_{2}e^{-x}-2ln(x)+e^x\Ei(-x)+e^{-x}\Ei(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=2ln(x)

y^{^{''}} - y = 2 ln (x)

y = c_{1} e^{x} + c_{2} e^{- x} - 2 ln (x) + e^{x} Ei (- x) + e^{- x} Ei (x)

Solution steps

y^{''} (x) - y = 2 ln (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} - 2 ln (x) + e^{x} Ei (- x) + e^{- x} Ei (x)

y = c_{1} e^{x} + c_{2} e^{- x} - 2 ln (x) + e^{x} Ei (- x) + e^{- x} Ei (x)

y^{^{''}} - 2 y^{^{'}} + y = 25 cos (2 x)

\frac{d ^{2} y}{d x ^{2}} = a y + b

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 t)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + 9 y = - 15 x + 7 + e^{x}

What is the solution for y^{''}-y=2ln(x) ?
The solution for y^{''}-y=2ln(x) is y=c_{1}e^x+c_{2}e^{-x}-2ln(x)+e^x\Ei(-x)+e^{-x}\Ei(x)