What is the solution for xy^'+(1+2x)y=10e^{-2x} ?

The solution for xy^'+(1+2x)y=10e^{-2x} is y=(10)/(e^{2x)}+(c_{1})/(e^{2x)x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+(1+2x)y=10e^{-2x}

x y^{^{'}} + (1 + 2 x) y = 10 e^{- 2 x}

y = \frac{10}{e ^{2 x}} + \frac{c _{1}}{e ^{2 x} x}

Solution steps

x y^{'} (x) + (1 + 2 x) y = 10 e^{- 2 x}

Solve linear ODE:

y = \frac{10}{e ^{2 x}} + \frac{c _{1}}{e ^{2 x} x}

y = \frac{10}{e ^{2 x}} + \frac{c _{1}}{e ^{2 x} x}

x \to + 1 lim (+ (\frac{1}{x ^{2} - 1}))

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

\frac{x}{( 2 x + 3 ) ^{3}}

\frac{\partial}{\partial x} (10 x^{0.5})

\frac{4}{s ^{2}} - \frac{10}{s}

What is the solution for xy^'+(1+2x)y=10e^{-2x} ?
The solution for xy^'+(1+2x)y=10e^{-2x} is y=(10)/(e^{2x)}+(c_{1})/(e^{2x)x}