What is the (y^3-x^3)y^'(x)=3x^2y+1,y(-2)=-1 ?

The (y^3-x^3)y^'(x)=3x^2y+1,y(-2)=-1 is -3x^2+1

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(y^3-x^3)y^'(x)=3x^2y+1,y(-2)=-1

(y^{3} - x^{3}) y^{^{'}} (x) = 3 x^{2} y + 1, y (- 2) = - 1

- 3 x^{2} + 1

Solution steps

Find

3 x^{2} y + 1

given

y (- 2) = - 1

3 x^{2} y + 1 = 3 x^{2} y + 1

Constant function results in a constant

y (- 2) = - 1

Expand

3 x^{2} (- 1) + 1 : - 3 x^{2} + 1

= - 3 x^{2} + 1

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What is the (y^3-x^3)y^'(x)=3x^2y+1,y(-2)=-1 ?
The (y^3-x^3)y^'(x)=3x^2y+1,y(-2)=-1 is -3x^2+1