en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-y^'+y=10

Solution

- y^{^{'}} + y = 10

Solution

y = - e^{t + c_{1}} + 10

Solution steps

- y^{'} (t) + y = 10

Solve separable ODE:

y = - e^{t + c_{1}} + 10

y = - e^{t + c_{1}} + 10

Graph

Popular Examples

(dy)/(dx)=-3/9 x^5y^8

\frac{d y}{d x} = - \frac{3}{9} x^{5} y^{8}

e^yy^'=((ln(t))^2)/t ,y(1)=ln(15)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (15)

\frac{d y}{d x} = - \frac{y}{a}

(dv)/(dt)=5-0.5*v

\frac{d v}{d t} = 5 - 0.5 \cdot v

y^'+4y=3,y(0)=0

y^{^{'}} + 4 y = 3, y (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for -y^'+y=10 ?
The solution for -y^'+y=10 is y=-e^{t+c_{1}}+10