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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{4 e ^{x} x ^{\frac{5}{2}}}{15}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{4 e ^{x} x ^{\frac{5}{2}}}{15}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{4 e ^{x} x ^{\frac{5}{2}}}{15}

9 y^{^{''}} - 16 y = sin (x)

y^{^{'''}} + y^{^{'}} - 10 y = 29 e^{4 x}

f^{^{''}} (x) + 3 f^{^{'}} (x) + 2 f (x) = 2, f (0) = 3, f^{^{'}} (0) = 5

y^{^{''}} - 6 y^{^{'}} + 10 y = 180 sin (2 x)

y^{^{''}} - 9 y^{^{'}} + 20 y = e^{x} (- 9 - 78 x + 36 x^{2})

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