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Popular >

y^'+2y=1,y(0)= 5/2

Solution

y^{^{'}} + 2 y = 1, y (0) = \frac{5}{2}

Solution

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

Solution steps

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

Solve linear ODE:

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

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

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\int x (2 x + 3)^{2} d x

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

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\frac{d}{d x} (12)

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\frac{d}{d x} (x ln (cos (x)))

limit as x approaches+0 of (2(y+x)-2y)/x

x \to + 0 lim (\frac{2 ( y + x ) - 2 y}{x})

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y=1,y(0)= 5/2 ?
The solution for y^'+2y=1,y(0)= 5/2 is y= 1/2+2/(e^{2t)}