What is the x(3y^2-2)y^'(x)-y^3=0,y(1)=2 ?

The x(3y^2-2)y^'(x)-y^3=0,y(1)=2 is 3xy'(x)y^2-2xy'(x)-16\circ

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x(3y^2-2)y^'(x)-y^3=0,y(1)=2

x (3 y^{2} - 2) y^{^{'}} (x) - y^{3} = 0, y (1) = 2

3 x y' (x) y^{2} - 2 x y' (x) - 16 \circ

Solution steps

Find

x (3 y^{2} - 2) y^{'} (x) - y^{3} \circ y

given

x (3 y^{2} - 2) y^{'} (x) - y^{3} = 0, y (1) = 2

x (3 y^{2} - 2) y^{^{'}} (x) - y^{3} \circ y = x (3 y^{2} - 2) y^{^{'}} (x) - y^{3} \circ y

y^{2} : 2^{2}

y^{3} : 2^{3}

y : 2

= x (3 y^{2} - 2) y^{^{'}} (x) - 2^{3} \circ 2

Expand

x (3 y^{2} - 2) y^{'} (x) - 2^{3} \circ 2 : 3 x y^{'} (x) y^{2} - 2 x y^{'} (x) - 16 \circ

= 3 x y^{'} (x) y^{2} - 2 x y^{'} (x) - 16 \circ

What is the x(3y^2-2)y^'(x)-y^3=0,y(1)=2 ?
The x(3y^2-2)y^'(x)-y^3=0,y(1)=2 is 3xy'(x)y^2-2xy'(x)-16\circ