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Popular >

y^'+2y=-5

Solution

y^{^{'}} + 2 y = - 5

Solution

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{5}{2}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{5}{2}

y = \frac{e ^{- 2 t - c_{1}}}{2} - \frac{5}{2}

Popular Examples

3 x^{^{'}} + x = 1

y^'=(8cos(8x))/(3+2y),y(0)=-1

y^{^{'}} = \frac{8 cos ( 8 x )}{3 + 2 y}, y (0) = - 1

t^9(dy)/(dt)+2t^8y=0

t^{9} \frac{d y}{d t} + 2 t^{8} y = 0

(dy)/(dx)=-1*1/((x+9)^2)*(y+1)

\frac{d y}{d x} = - 1 \cdot \frac{1}{( x + 9 ) ^{2}} \cdot (y + 1)

(dy)/(dx)=4+y^2

\frac{d y}{d x} = 4 + y^{2}

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y=-5 ?
The solution for y^'+2y=-5 is y=(e^{-2t-c_{1}})/2-5/2