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

解

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})

解答ステップ

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

分離可能 ODE を解く:

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})