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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{9 x}{x ^{3} + 1} d x

\frac{\partial}{\partial x} (2 - (x^{2} + y^{2})^{\frac{1}{7}})

y = (x^{5} - 1)^{3}

x \to 0 lim (x sin (\frac{1}{∣ x ∣}))

- 2 e^{- x^{2}} x

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