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Popular Calculus >

y^'+2y=0,y(0)=0

Solution

y^{^{'}} + 2 y = 0, y (0) = 0

Solution

y = 0

Solution steps

y^{'} (t) + 2 y = 0

Solve linear ODE:

y = 0

y = 0

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+2y=0,y(0)=0 ?
The solution for y^'+2y=0,y(0)=0 is y=0