y^'=-xy+4x

Lösung

y^{^{'}} = - x y + 4 x

y = e^{- \frac{x ^{2}}{2} + c_{1}} + 4

Schritte zur Lösung

y^{'} (x) = - x y + 4 x

Solve separable ODE:

y = e^{- \frac{x ^{2}}{2} + c_{1}} + 4

y = e^{- \frac{x ^{2}}{2} + c_{1}} + 4

\frac{d}{d x} (tan (y)) = x

(x y^{2} + 3 y^{2}) d y - 9 y d x = 0

x d y = (\frac{1}{y ^{2}} - y) d x

\frac{d y}{d x} = \frac{3 x ^{2} - e ^{x}}{2 y - 5}, y (0) = 1

y^{^{'}} = y^{2} \cdot x