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Popular Calculus >

y^'+8y=1,y(0)=1

Solution

y^{^{'}} + 8 y = 1, y (0) = 1

Solution

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

Solution steps

y^{'} (t) + 8 y = 1

Solve linear ODE:

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+8y=1,y(0)=1 ?
The solution for y^'+8y=1,y(0)=1 is y= 1/8+7/(8e^{8t)}