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Popular >

y^'+8y=4,y(0)=3

Solution

y^{^{'}} + 8 y = 4, y (0) = 3

Solution

y = \frac{1}{2} + \frac{5}{2 e ^{8 t}}

Solution steps

y^{'} (t) + 8 y = 4

Solve linear ODE:

y = \frac{1}{2} + \frac{5}{2 e ^{8 t}}

y = \frac{1}{2} + \frac{5}{2 e ^{8 t}}

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limit as x approaches 1 of (2x)/(x^3+1)

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Frequently Asked Questions (FAQ)

What is the solution for y^'+8y=4,y(0)=3 ?
The solution for y^'+8y=4,y(0)=3 is y= 1/2+5/(2e^{8t)}