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Popular >

y^'+4y=2

Solution

y^{^{'}} + 4 y = 2

Solution

y = - \frac{e ^{- 4 t - c_{1}}}{2} + \frac{1}{2}

Solution steps

y^{'} (t) + 4 y = 2

Solve separable ODE:

y = - \frac{e ^{- 4 t - c_{1}}}{2} + \frac{1}{2}

y = - \frac{e ^{- 4 t - c_{1}}}{2} + \frac{1}{2}

Popular Examples

(dy)/(dx)=e^xcos^2(y)

\frac{d y}{d x} = e^{x} cos^{2} (y)

(x^2+1)(dy)/(dx)+3x(y-1)=0,y(0)=2

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 2

y^'sec(t)=(-y^{17})/(16)

y^{^{'}} sec (t) = \frac{- y ^{17}}{16}

(x^2+1)(dy)/(dx)+3xy=3x,y(0)=2

(x^{2} + 1) \frac{d y}{d x} + 3 x y = 3 x, y (0) = 2

(x+2)sin(y)+xcos(y)(dy)/(dx)=0

(x + 2) sin (y) + x cos (y) \frac{d y}{d x} = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y=2 ?
The solution for y^'+4y=2 is y=-(e^{-4t-c_{1}})/2+1/2