What is the solution for y^{'''}-2y^{''}+y^'=-2e^x-3x ?

The solution for y^{'''}-2y^{''}+y^'=-2e^x-3x is y=c_{1}+c_{2}e^x+c_{3}xe^x-e^xx^2-(3x^2)/2-6x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + c_{2} e^{x} + c_{3} x e^{x} - e^{x} x^{2} - \frac{3 x ^{2}}{2} - 6 x

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{x} + c_{3} x e^{x} - e^{x} x^{2} - \frac{3 x ^{2}}{2} - 6 x

y = c_{1} + c_{2} e^{x} + c_{3} x e^{x} - e^{x} x^{2} - \frac{3 x ^{2}}{2} - 6 x

y^{^{'''}} - 6 y^{^{''}} = x^{2}

y^{^{'''}} - 6 y^{^{'}} = 3 - cos (x)

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

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

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} csc (x)

What is the solution for y^{'''}-2y^{''}+y^'=-2e^x-3x ?
The solution for y^{'''}-2y^{''}+y^'=-2e^x-3x is y=c_{1}+c_{2}e^x+c_{3}xe^x-e^xx^2-(3x^2)/2-6x