y^'=((3x^2-e^x))/(2y-5),y(0)=1

Lösung

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

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

y^{^{'}} = \frac{y - a}{b - x}

\frac{d y}{d x} = x - x y - y + 1

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

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

d x + sin (t) x d t sin (t) d t = 0