What is the solution for y^{'''}-2y^{''}+y^'=1 ?

The solution for y^{'''}-2y^{''}+y^'=1 is y=c_{1}+c_{2}e^t+c_{3}te^t+t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}-2y^{''}+y^'=1

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t

Solution steps

y^{'''} (t) - 2 y^{''} (t) + y^{'} (t) = 1

Solve linear ODE:

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t

\frac{1}{6} y^{^{''}} + 6 y = tan (6 x) - \frac{1}{3} e^{3 x}

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

y^{^{'''}} - y^{^{'}} = 10 cos (2 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{3} - 1

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

What is the solution for y^{'''}-2y^{''}+y^'=1 ?
The solution for y^{'''}-2y^{''}+y^'=1 is y=c_{1}+c_{2}e^t+c_{3}te^t+t