What is the solution for y^'+4y=e^{-x},y(0)=10 ?

The solution for y^'+4y=e^{-x},y(0)=10 is y= 1/(3e^x)+(29)/(3e^{4x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4y=e^{-x},y(0)=10

y^{^{'}} + 4 y = e^{- x}, y (0) = 10

y = \frac{1}{3 e ^{x}} + \frac{29}{3 e ^{4 x}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{3 e ^{x}} + \frac{29}{3 e ^{4 x}}

y = \frac{1}{3 e ^{x}} + \frac{29}{3 e ^{4 x}}

\int_{0}^{6} \frac{1}{∣ x - 3 ∣} d x

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

y^{^{''}} + 9 y = 2 x^{2} - 5

x \to 1 lim (\frac{x ^{10} - 1}{x - 1})

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

What is the solution for y^'+4y=e^{-x},y(0)=10 ?
The solution for y^'+4y=e^{-x},y(0)=10 is y= 1/(3e^x)+(29)/(3e^{4x)}