What is the solution for ((y^'-e-t+4))/y =-4 ?

The solution for ((y^'-e-t+4))/y =-4 is y= t/4+e/4-17/16+(c_{1})/(e^{4t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

((y^'-e-t+4))/y =-4

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

y = \frac{t}{4} + \frac{e}{4} - \frac{17}{16} + \frac{c _{1}}{e ^{4 t}}

Solution steps

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

Solve linear ODE:

y = \frac{t}{4} + \frac{e}{4} - \frac{17}{16} + \frac{c _{1}}{e ^{4 t}}

y = \frac{t}{4} + \frac{e}{4} - \frac{17}{16} + \frac{c _{1}}{e ^{4 t}}

x \to 0 lim (\frac{7 ^{s i n (x)} - 1}{x})

\int \frac{1}{9} x^{- \frac{8}{9}} d x

\frac{\partial}{\partial x} (x^{3} e^{y} sin (x - y))

t an g e n t f (x) = x^{2} - 3, (3, 6)

ma c l a u r in \frac{4 x}{x ^{2} + 2 x - 3}

What is the solution for ((y^'-e-t+4))/y =-4 ?
The solution for ((y^'-e-t+4))/y =-4 is y= t/4+e/4-17/16+(c_{1})/(e^{4t)}