sqrt(x^2+6)dx-1/(xcsc(-8y))dy=0

Lösung

x^{2} + 6 d x - \frac{1}{x csc ( - 8 y )} d y = 0

y = \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}, y = - \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}

Schritte zur Lösung

x^{2} + 6 d x - \frac{1}{x csc ( - 8 y )} d y = 0

Solve separable ODE:

y = \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}, y = - \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}

y = \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}, y = - \frac{arccos ( \frac{8}{3} ( x ^{2} + 6 ) ^{\frac{3}{2}} + c _{1} )}{8} + \frac{πn}{4}

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

x \cdot \frac{d x}{d t} = e^{3 x} t

x^{^{'}} = e^{x} \cdot t \cdot cos (t)

\frac{d P}{d t} = 4 Pt, P (1) = 6

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