What is the solution for (1-x^2)y^'=6xy,y(0)=3 ?

The solution for (1-x^2)y^'=6xy,y(0)=3 is y= 3/((1-x^2)^3)

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(1-x^2)y^'=6xy,y(0)=3

(1 - x^{2}) y^{^{'}} = 6 x y, y (0) = 3

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

Solution steps

(1 - x^{2}) y^{'} (x) = 6 x y

Solve separable ODE:

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

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

3 e^{2 x} \frac{d y}{d x} = - 4 \cdot \frac{x}{y ^{2}}

\frac{d y}{d x} = \frac{( 1 - y )}{xv}

y^{^{'}} = 1 - \frac{y}{10}

4 \frac{d y}{d t} - \frac{1}{2} \frac{d y}{d t} = a + 32 - 4 y

20 \frac{d y}{d x} + y = 1

What is the solution for (1-x^2)y^'=6xy,y(0)=3 ?
The solution for (1-x^2)y^'=6xy,y(0)=3 is y= 3/((1-x^2)^3)