What is the solution for y^{'''}+y^{''}=4x^2+3e^x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}+y^{''}=4x^2+3e^x

y^{^{'''}} + y^{^{''}} = 4 x^{2} + 3 e^{x}

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

Solution steps

y^{'''} (x) + y^{''} (x) = 4 x^{2} + 3 e^{x}

Solve linear ODE:

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

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

y^{^{''''}} + 2 y^{^{''}} + y = 3 x + 4

y^{^{''}} - 18 y^{^{'}} + 81 y = 480 sin (x) - 210 cos (x)

y^{^{''}} + 9 y = 2, y^{^{'}} (0) = 5

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

y^{^{''}} - y^{^{'}} - 6 y = 250 t e^{- 2 t}

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