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Popular >

y^'+0.5y=3

Solution

y^{^{'}} + 0.5 y = 3

Solution

y = - 2 e^{\frac{- t - c _{1}}{2}} + 6

Solution steps

y^{'} (t) + 0.5 y = 3

Solve separable ODE:

y = - 2 e^{\frac{- t - c _{1}}{2}} + 6

y = - 2 e^{\frac{- t - c _{1}}{2}} + 6

Popular Examples

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

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

x^'=2+x+1/4 x-1/2

x^{^{'}} = 2 + x + \frac{1}{4} x - \frac{1}{2}

e^y*y^'=(3x^2)/(1+y)

e^{y} \cdot y^{^{'}} = \frac{3 x ^{2}}{1 + y}

(dx)/(dt)=-x/(100)+125

\frac{d x}{d t} = - \frac{x}{100} + 125

x^'=-x^2,x(0)=5

x^{^{'}} = - x^{2}, x (0) = 5

Frequently Asked Questions (FAQ)

What is the solution for y^'+0.5y=3 ?
The solution for y^'+0.5y=3 is y=-2e^{(-t-c_{1})/2}+6