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Popular >

2y^'+4y=1

Solution

2 y^{^{'}} + 4 y = 1

Solution

y = - \frac{e ^{- 2 t - c_{1}}}{4} + \frac{1}{4}

Solution steps

2 y^{'} (t) + 4 y = 1

Solve separable ODE:

y = - \frac{e ^{- 2 t - c_{1}}}{4} + \frac{1}{4}

y = - \frac{e ^{- 2 t - c_{1}}}{4} + \frac{1}{4}

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Frequently Asked Questions (FAQ)

What is the solution for 2y^'+4y=1 ?
The solution for 2y^'+4y=1 is y=-(e^{-2t-c_{1}})/4+1/4