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Popular >

y^'+1/2 y=0

Solution

y^{^{'}} + \frac{1}{2} y = 0

Solution

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

Solution steps

y^{'} (t) + \frac{1}{2} y = 0

Solve separable ODE:

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

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

Popular Examples

y^{''}+10y^'+y=0

y^{^{''}} + 10 y^{^{'}} + y = 0

derivative of ln^2(x+1)

\frac{d}{d x} (ln^{2} (x + 1))

derivative of (x^2+x^{1/2})

\frac{d}{d x} ((x^{2} + x)^{\frac{1}{2}})

derivative of ((x^3-1/(2x^3+1))^4)

\frac{d}{d x} ((\frac{x ^{3} - 1}{2 x ^{3} + 1})^{4})

(\partial)/(\partial x)(2^{xy})

\frac{\partial}{\partial x} (2^{x y})

Frequently Asked Questions (FAQ)

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