y^'=(3xsqrt(16+y^2))/y

Lösung

y^{^{'}} = \frac{3 x 16 + y ^{2}}{y}

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}

Schritte zur Lösung

y^{'} (x) = \frac{3 x 16 + y ^{2}}{y}

Solve separable ODE:

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}

y^{^{'}} = \frac{6 sin ( x )}{sin ( y )}

(\frac{d x}{d t}) = - x

2 v^{2} y^{2} d y + 2 v y^{3} d v + v^{2} y^{2} d y + y^{2} d y = 0

\frac{d v}{d t} = 4 - 7 v

e^{x} \frac{d y}{d x} + y^{3} = x y^{3}, y = 1, x = 0