What is the solution for (dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1 ?

The solution for (dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1 is y=-sqrt((ln(2e^{x^2)+e^3-2e))/3}

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(dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1

\frac{d y}{d x} = \frac{2 x}{3 y} e^{x^{2} - 3 y^{2}}, y (1) = - 1

y = - \frac{ln ( 2 e ^{x^{2}} + e ^{3} - 2 e )}{3}

Solution steps

\frac{d y}{d x} = \frac{2 x}{3 y} e^{x^{2} - 3 y^{2}}

Solve separable ODE:

y = - \frac{ln ( 2 e ^{x^{2}} + e ^{3} - 2 e )}{3}

y = - \frac{ln ( 2 e ^{x^{2}} + e ^{3} - 2 e )}{3}

\frac{( x ^{2} - 9 )}{x} \frac{d y}{d x} + y = 0

x \frac{d y}{d x} = e^{5 y + x}

y^{^{'}} = \frac{x + x y ^{2}}{4 y}, y (1) = 0

y^{^{'}} - 9 x^{2} y + y = 0

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

What is the solution for (dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1 ?
The solution for (dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1 is y=-sqrt((ln(2e^{x^2)+e^3-2e))/3}