(3x^2+2)^3y^'-xy^2=0,y(0)= 3/2

Lösung

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

y = \frac{432 x ^{4}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{576 x ^{2}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{192}{279 x ^{4} + 372 x ^{2} + 128}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{432 x ^{4}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{576 x ^{2}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{192}{279 x ^{4} + 372 x ^{2} + 128}

y = \frac{432 x ^{4}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{576 x ^{2}}{279 x ^{4} + 372 x ^{2} + 128} + \frac{192}{279 x ^{4} + 372 x ^{2} + 128}

\frac{d r}{d t} = 0.1872 r

x tan (x) d x - y cos (x) d y = 0

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

\frac{d y}{d t} = y t^{3} - 1.5 y

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