en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=t-y

Solution

y^{^{'}} = t - y

Solution

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

Solution steps

y^{'} (t) = t - y

Solve linear ODE:

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

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

Graph

Popular Examples

area xcos(x),-pi/2 , pi/2 area

x cos (x), - \frac{π}{2}, \frac{π}{2}

(\partial)/(\partial x)(-cos(x)+xcos(y))

\frac{\partial}{\partial x} (- cos (x) + x cos (y))

derivative of xcos(xcos(y))

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

integral of e^{x/2}

\int e^{\frac{x}{2}} d x

derivative of 7x^{-12}

\frac{d}{d x} (7 x^{- 12})

Frequently Asked Questions (FAQ)

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