en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^'+y=5,y(3)=0

Solution

2 y^{^{'}} + y = 5, y (3) = 0

Solution

y = - 5 e^{\frac{- t + 3}{2}} + 5

Solution steps

2 y^{'} (t) + y = 5

Solve separable ODE:

y = - 5 e^{\frac{- t + 3}{2}} + 5

y = - 5 e^{\frac{- t + 3}{2}} + 5

Popular Examples

xy^2(dy)/(dx)=x+6

x y^{2} \frac{d y}{d x} = x + 6

y^'-4y=6,y(0)=0

y^{^{'}} - 4 y = 6, y (0) = 0

(dy)/(dx)+5y=-4

\frac{d y}{d x} + 5 y = - 4

y^'=(-(3y^2+1)x)/(x^2+1)

y^{^{'}} = \frac{- ( 3 y ^{2} + 1 ) x}{x ^{2} + 1}

derivative of 7x^2+8ln(xy)=14

\frac{d}{d x} (7 x^{2} + 8 ln (x y)) = 14

Frequently Asked Questions (FAQ)

What is the solution for 2y^'+y=5,y(3)=0 ?
The solution for 2y^'+y=5,y(3)=0 is y=-5e^{(-t+3)/2}+5