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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

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