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Popular >

y^'-(2x+1)y=0

Solution

y^{^{'}} - (2 x + 1) y = 0

Solution

y = e^{x^{2} + x + c_{1}}

Solution steps

y^{'} (x) - (2 x + 1) y = 0

Solve separable ODE:

y = e^{x^{2} + x + c_{1}}

y = e^{x^{2} + x + c_{1}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'-(2x+1)y=0 ?
The solution for y^'-(2x+1)y=0 is y=e^{x^2+x+c_{1}}