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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{x ^{3}}{3 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{3}}{3 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

y = \frac{x ^{3}}{3 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

3 t + 1

(sin (4 x))^{^{'}}

\int_{0}^{b} \frac{1}{9 - x ^{2}} d x

\int - 9.8 x d x

(4 t + 2 y + 5) d t - d y = 0

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