What is the solution for y^{''}-2y^'+y=15sqrt(x)e^x ?

The solution for y^{''}-2y^'+y=15sqrt(x)e^x is y=c_{1}e^x+c_{2}xe^x+4e^xx^{5/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=15sqrt(x)e^x

y^{^{''}} - 2 y^{^{'}} + y = 15 x e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + 4 e^{x} x^{\frac{5}{2}}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 15 x e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 e^{x} x^{\frac{5}{2}}

y = c_{1} e^{x} + c_{2} x e^{x} + 4 e^{x} x^{\frac{5}{2}}

y^{^{''}} + 2 y^{^{'}} + y = ln (x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 5 e^{2} x

a y^{^{''}} + b y^{^{'}} + y = c

y^{^{''}} + 4 y^{^{'}} - 5 y = 3 e^{t}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{- 4} \cdot e^{2 x}

What is the solution for y^{''}-2y^'+y=15sqrt(x)e^x ?
The solution for y^{''}-2y^'+y=15sqrt(x)e^x is y=c_{1}e^x+c_{2}xe^x+4e^xx^{5/2}