y^'=(3x-e^x)/(2y-5)

Lösung

y^{^{'}} = \frac{3 x - e ^{x}}{2 y - 5}

y = \frac{5 + - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}, y = \frac{5 - - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}

Schritte zur Lösung

y^{'} (x) = \frac{3 x - e ^{x}}{2 y - 5}

Solve separable ODE:

y = \frac{5 + - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}, y = \frac{5 - - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}

y = \frac{5 + - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}, y = \frac{5 - - 2 ( - 3 x ^{2} + 2 e ^{x} - c _{1} ) + 25}{2}

(1 + y^{2}) cos (x) d x = 2 (1 + sin^{2} (x)) y d y

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

y^{^{'}} + (1 - n^{2}) y = \frac{2 ( - 1 ) ^{n}}{n}

y^{^{'}} = \frac{y ^{2}}{x} + \frac{1}{x}

\frac{d y}{d x} y = x^{2} - 4 x