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

The solution for y^'=(3x)/(y+x^2y),y(0)=-3 is y=-sqrt(3ln(1+x^2)+9)

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

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

y = - 3 ln (1 + x^{2}) + 9

Solution steps

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

Solve separable ODE:

y = - 3 ln (1 + x^{2}) + 9

y = - 3 ln (1 + x^{2}) + 9

\frac{\partial}{\partial z} (\frac{1}{z})

\int \frac{w ^{2} + 16}{w ^{2} - 16} d w

\frac{1}{( s - 2 ) ( s - 4 )}

y^{^{'''}} - 12 y^{^{''}} + 52 y^{^{'}} - 96 y = 0

What is the solution for y^'=(3x)/(y+x^2y),y(0)=-3 ?
The solution for y^'=(3x)/(y+x^2y),y(0)=-3 is y=-sqrt(3ln(1+x^2)+9)