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Popular >

y^'+y=-2

Solution

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

Solution

y = e^{- t - c_{1}} - 2

Solution steps

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

Solve separable ODE:

y = e^{- t - c_{1}} - 2

y = e^{- t - c_{1}} - 2

Popular Examples

sqrt(x)(dy)/(dx)=e^{10y+sqrt(x)},x>0

x \frac{d y}{d x} = e^{10 y + x}, x > 0

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

y^'=sqrt(1-y^2),y(0)=0

y^{^{'}} = 1 - y^{2}, y (0) = 0

x/(a^2)+(ydy)/(b^2dx)=0

\frac{x}{a ^{2}} + \frac{y d y}{b ^{2} d x} = 0

cos(y)y^'+2xsin(y)=-2x

cos (y) y^{^{'}} + 2 x sin (y) = - 2 x

Frequently Asked Questions (FAQ)

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