What is the solution for y^{'''}-y^'=2sin(3x) ?

The solution for y^{'''}-y^'=2sin(3x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+1/15 cos(3x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}-y^'=2sin(3x)

y^{^{'''}} - y^{^{'}} = 2 sin (3 x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

Solution steps

y^{'''} (x) - y^{'} (x) = 2 sin (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

\int \frac{3}{64} x^{2} d x

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

x \to 0 lim (\frac{sin ( x ^{9} )}{x})

\frac{\partial}{\partial x} (\frac{x - b}{b - c})

\frac{\partial}{\partial x} (x^{2} y + x sin (y))

What is the solution for y^{'''}-y^'=2sin(3x) ?
The solution for y^{'''}-y^'=2sin(3x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+1/15 cos(3x)