y^'=(sin^5(x)cos^2(x))(y^2+1),y(0)=0

Lösung

y^{^{'}} = (sin^{5} (x) cos^{2} (x)) (y^{2} + 1), y (0) = 0

y = tan (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7} + \frac{8}{105})

Schritte zur Lösung

y^{'} (x) = (sin^{5} (x) cos^{2} (x)) (y^{2} + 1)

Solve separable ODE:

y = tan (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7} + \frac{8}{105})

y = tan (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7} + \frac{8}{105})

3 e^{x} \cdot tan (y) d x + (2 - e^{x}) sec^{2} (y) d y = 0

\frac{d v}{d t} = 3 - 4 v

y^{^{'}} = y (5 - y) - \frac{25}{4}, y (0) = a

a^{2} (d y - d x) = x^{2} d y + y^{2} d x

y \frac{d y}{d x} = \frac{1}{1 - y ^{2}}