What is the solution for (y^'-e^{-t}+3)/y =-3,y(0)=4 ?

The solution for (y^'-e^{-t}+3)/y =-3,y(0)=4 is y= 1/(2e^t)-1+9/(2e^{3t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y^'-e^{-t}+3)/y =-3,y(0)=4

\frac{y ^{^{'}} - e ^{- t} + 3}{y} = - 3, y (0) = 4

y = \frac{1}{2 e ^{t}} - 1 + \frac{9}{2 e ^{3 t}}

Solution steps

\frac{y ^{'} ( t ) - e ^{- t} + 3}{y} = - 3

Solve linear ODE:

y = \frac{1}{2 e ^{t}} - 1 + \frac{9}{2 e ^{3 t}}

y = \frac{1}{2 e ^{t}} - 1 + \frac{9}{2 e ^{3 t}}

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (t)

\frac{d}{d x} (5 x cos (x))

x \to 0 + lim (3 - x)

\frac{d}{d x} (3 sin (x) + 3 cos (x))

\int ln (x + 16) d x

What is the solution for (y^'-e^{-t}+3)/y =-3,y(0)=4 ?
The solution for (y^'-e^{-t}+3)/y =-3,y(0)=4 is y= 1/(2e^t)-1+9/(2e^{3t)}