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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=x^{-1/2}e^x

y^{^{''}} - 2 y^{^{'}} + y = x^{- \frac{1}{2}} e^{x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x^{- \frac{1}{2}} e^{x}

Solve linear ODE:

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

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

1000 - 0.017 y^{^{'}} = 100 y^{^{''}}

y^{^{''}} + 8 y^{^{'}} + 3 y = 2 x^{2} + 3

y^{^{''}} - 4 y = 2 x^{- 1} e^{3 x}

x^{^{''}} - 3 x^{^{'}} = 8 sin (t)

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

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