What is the solution for 6y^'-3y=2e^{-2x},y(0)=10 ?

The solution for 6y^'-3y=2e^{-2x},y(0)=10 is y=-2/(15e^{2x)}+152/15 e^{x/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6y^'-3y=2e^{-2x},y(0)=10

6 y^{^{'}} - 3 y = 2 e^{- 2 x}, y (0) = 10

y = - \frac{2}{15 e ^{2 x}} + \frac{152}{15} e^{\frac{x}{2}}

Solution steps

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

Solve linear ODE:

y = - \frac{2}{15 e ^{2 x}} + \frac{152}{15} e^{\frac{x}{2}}

y = - \frac{2}{15 e ^{2 x}} + \frac{152}{15} e^{\frac{x}{2}}

ln (1 + 2 x), 1

\int cos^{3} (\frac{x}{3}) d x

y = x + 2, y = \frac{1}{x + 1}, x = 0, x = 2

\frac{d y}{d t} = y^{1.01}

\int \frac{ln ( x )}{x ^{1.2}} d x

What is the solution for 6y^'-3y=2e^{-2x},y(0)=10 ?
The solution for 6y^'-3y=2e^{-2x},y(0)=10 is y=-2/(15e^{2x)}+152/15 e^{x/2}