What is the solution for y^'=(15x^2+4)(6y+5) ?

The solution for y^'=(15x^2+4)(6y+5) is y=(e^{30x^3+24x+c_{1}})/6-5/6

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y^'=(15x^2+4)(6y+5)

y^{^{'}} = (15 x^{2} + 4) (6 y + 5)

y = \frac{e ^{30 x^{3} + 24 x + c_{1}}}{6} - \frac{5}{6}

Solution steps

y^{'} (x) = (15 x^{2} + 4) (6 y + 5)

Solve separable ODE:

y = \frac{e ^{30 x^{3} + 24 x + c_{1}}}{6} - \frac{5}{6}

y = \frac{e ^{30 x^{3} + 24 x + c_{1}}}{6} - \frac{5}{6}

x tan (y) + (1 + x^{2}) \frac{d y}{d x} = 0

\frac{d P}{d t} = a P (t) - b P (t)^{2}

y - 2 y^{^{'}} + 10 y = 0

\frac{d v}{d t} = 10 - \frac{1}{100} v, v = 0, t = 0

\frac{d y}{d x} = \frac{x 4 - y ^{2}}{1 + x ^{2}}

What is the solution for y^'=(15x^2+4)(6y+5) ?
The solution for y^'=(15x^2+4)(6y+5) is y=(e^{30x^3+24x+c_{1}})/6-5/6