cos^3(y)(dy)/(dx)=xcsc(y),y(0)= pi/3

Lösung

cos^{3} (y) \frac{d y}{d x} = x csc (y), y (0) = \frac{π}{3}

y = arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn

Schritte zur Lösung

cos^{3} (y) \frac{d y}{d x} = x csc (y)

Solve Bernoulli ODE:

y = arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn

y = arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (\frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn, y = - arccos (- \frac{4 - 32 x ^{2} + 1}{2}) + 2 πn

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

\frac{d v}{d t} = 4 - 8 v

\frac{d y}{d t} = \frac{16 t ^{3}}{y}

f^{^{'}} (x) + a \cdot f (x) = a \cdot b

y^{^{'}} + 2 y + 6 = 0