What is the solution for y^'=(4t^3+1)/(2y-6),y(1)=2 ?

The solution for y^'=(4t^3+1)/(2y-6),y(1)=2 is y=3-sqrt(t^4+t-1)

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y^'=(4t^3+1)/(2y-6),y(1)=2

y^{^{'}} = \frac{4 t ^{3} + 1}{2 y - 6}, y (1) = 2

y = 3 - t^{4} + t - 1

Solution steps

y^{'} (t) = \frac{4 t ^{3} + 1}{2 y - 6}

Solve separable ODE:

y = 3 - t^{4} + t - 1

y = 3 - t^{4} + t - 1

\int \frac{e ^{\frac{6}{t}}}{t ^{2}} d t

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - 3 x z)

\frac{d}{d x} (x^{2} + x + 2)

\frac{d y}{d x}, 2 x^{2} + y^{2} - 7 = 0

What is the solution for y^'=(4t^3+1)/(2y-6),y(1)=2 ?
The solution for y^'=(4t^3+1)/(2y-6),y(1)=2 is y=3-sqrt(t^4+t-1)