What is the solution for y^'=(x^2+7x+3)/(y^2),y(0)=3 ?

The solution for y^'=(x^2+7x+3)/(y^2),y(0)=3 is y=\sqrt[3]{x^3+(21x^2)/2+9x+27}

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y^'=(x^2+7x+3)/(y^2),y(0)=3

y^{^{'}} = \frac{x ^{2} + 7 x + 3}{y ^{2}}, y (0) = 3

Solution steps

y^{'} (x) = \frac{x ^{2} + 7 x + 3}{y ^{2}}

Solve separable ODE:

4 d y + y d x = x^{2} d y, y (4) = - 1

y^{^{'}} + (x + 2) y = 0

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{7} e^{- x} d y = 0

e^{x} \frac{d y}{d x} + y^{2} = x y^{2}

y \cdot d x = - (1 - x) d y

What is the solution for y^'=(x^2+7x+3)/(y^2),y(0)=3 ?
The solution for y^'=(x^2+7x+3)/(y^2),y(0)=3 is y=\sqrt[3]{x^3+(21x^2)/2+9x+27}