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Popular >

2y^'+y=1

Solution

2 y^{^{'}} + y = 1

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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\int \frac{x + 5}{2 x ^{3} - 8 x} d x

(\partial)/(\partial x)(cos(x+3y))

\frac{\partial}{\partial x} (cos (x + 3 y))

derivative of 3x^2-2x-1

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

Frequently Asked Questions (FAQ)

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