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Popular >

2x^'+x=0

Solution

2 x^{^{'}} + x = 0

Solution

x = e^{\frac{- t + c _{1}}{2}}

Solution steps

2 x^{'} (t) + x = 0

Solve separable ODE:

x = e^{\frac{- t + c _{1}}{2}}

x = e^{\frac{- t + c _{1}}{2}}

Popular Examples

(dy)/(dx)sin(2y)-cos(x)=0

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

(dr)/(dt)=0.1872r,r(0)=24

\frac{d r}{d t} = 0.1872 r, r (0) = 24

y^'+2xy^2=0,y(0)= 1/(x^2+c),y(0)=1

y^{^{'}} + 2 x y^{2} = 0, y (0) = \frac{1}{x ^{2} + c}, y (0) = 1

(x - y x) d y + y d x = 0

((dy))/((dx))=e^{(3y-x)}

\frac{( d y )}{( d x )} = e^{(3 y - x)}

Frequently Asked Questions (FAQ)

What is the solution for 2x^'+x=0 ?
The solution for 2x^'+x=0 is x=e^{(-t+c_{1})/2}