What is the solution for 2sin(y^2)+tycos(y^2)y^'=0 ?

The solution for 2sin(y^2)+tycos(y^2)y^'=0 is

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2sin(y^2)+tycos(y^2)y^'=0

2 sin (y^{2}) + t y cos (y^{2}) y^{^{'}} = 0

y = arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = - arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn, y = - π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn

Solution steps

2 sin (y^{2}) + t y cos (y^{2}) y^{'} (t) = 0

Solve separable ODE:

y = arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = - arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn, y = - π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn

y = arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = - arcsin (\frac{c _{1}}{t ^{4}}) + 2 πn, y = π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn, y = - π + arcsin (- \frac{c _{1}}{t ^{4}}) + 2 πn

x^{2} sin (3 y) + cos (3 y) \frac{d y}{d x} = 4 x^{2}

\frac{d y}{d x} - 5 \frac{d y}{d x} + 6 y = 0

\frac{d y}{d t} = y t^{2} - y

x y - x \frac{d y}{d x} = 2 x, y (- 1) = 3

\frac{d x}{d y} = 50 - \frac{x}{100}

What is the solution for 2sin(y^2)+tycos(y^2)y^'=0 ?
The solution for 2sin(y^2)+tycos(y^2)y^'=0 is