What is the solution for f^'(x)=2f(x)+4,f(0)=1 ?

The solution for f^'(x)=2f(x)+4,f(0)=1 is f(x)=3e^{2x}-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f^'(x)=2f(x)+4,f(0)=1

f^{^{'}} (x) = 2 f (x) + 4, f (0) = 1

f (x) = 3 e^{2 x} - 2

Solution steps

f^{'} (x) = 2 f (x) + 4

Solve separable ODE:

f (x) = 3 e^{2 x} - 2

f (x) = 3 e^{2 x} - 2

y^{^{'}} = - 0.3 y

\frac{d y}{d x} = x^{4} y^{- 7}, y (0) = 3

\frac{d y}{d x} = (\frac{1 + y}{2 + x})

y^{^{'}} = \frac{( 3 x ^{2} - e ^{x} )}{2 y - 5}, y (0) = 1

y^{^{'}} = \frac{y - a}{b - x}

What is the solution for f^'(x)=2f(x)+4,f(0)=1 ?
The solution for f^'(x)=2f(x)+4,f(0)=1 is f(x)=3e^{2x}-2