What is the solution for y^'-y=1+3sin(t) ?

The solution for y^'-y=1+3sin(t) is y=-1+(3e^t(-e^{-t}cos(t)-e^{-t}sin(t)))/2+c_{1}e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-y=1+3sin(t)

y^{^{'}} - y = 1 + 3 sin (t)

y = - 1 + \frac{3 e ^{t} ( - e ^{- t} cos ( t ) - e ^{- t} sin ( t ) )}{2} + c_{1} e^{t}

Solution steps

y^{'} (t) - y = 1 + 3 sin (t)

Solve linear ODE:

y = - 1 + \frac{3 e ^{t} ( - e ^{- t} cos ( t ) - e ^{- t} sin ( t ) )}{2} + c_{1} e^{t}

y = - 1 + \frac{3 e ^{t} ( - e ^{- t} cos ( t ) - e ^{- t} sin ( t ) )}{2} + c_{1} e^{t}

f (x) = 4 lo g_{6} (2)

f (x) = \frac{1}{2 x}, at x = 65

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

\int_{0}^{1} \frac{10 θ}{( 1 + θ ^{\frac{3}{2}} ) ^{2}} d θ

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

What is the solution for y^'-y=1+3sin(t) ?
The solution for y^'-y=1+3sin(t) is y=-1+(3e^t(-e^{-t}cos(t)-e^{-t}sin(t)))/2+c_{1}e^t