What is the solution for yy^'+6x=0,y^2(0)=-6x^2+c,y(0)=4 ?

The solution for yy^'+6x=0,y^2(0)=-6x^2+c,y(0)=4 is y=sqrt(-6x^2+c_{1)},y=-sqrt(-6x^2+c_{1)}

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yy^'+6x=0,y^2(0)=-6x^2+c,y(0)=4

y y^{^{'}} + 6 x = 0, y^{2} (0) = - 6 x^{2} + c, y (0) = 4

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

Solution steps

y y^{'} (x) + 6 x = 0

Solve separable ODE:

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

y^{^{'}} = - \frac{3}{7} e^{3 y}

x y^{^{'}} - 3 y + 2 = 0

\frac{d y}{d x} = 6 y ln (\frac{2}{y})

- \frac{d h}{d t} = 5 h

ln (x) y^{^{'}} - \frac{2 y}{x} = 0

What is the solution for yy^'+6x=0,y^2(0)=-6x^2+c,y(0)=4 ?
The solution for yy^'+6x=0,y^2(0)=-6x^2+c,y(0)=4 is y=sqrt(-6x^2+c_{1)},y=-sqrt(-6x^2+c_{1)}