What is the solution for y^'=(xy^3)/4 ,y(0)=6 ?

The solution for y^'=(xy^3)/4 ,y(0)=6 is y=sqrt(-(36)/(9x^2-1))

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y^'=(xy^3)/4 ,y(0)=6

y^{^{'}} = \frac{x y ^{3}}{4}, y (0) = 6

y = - \frac{36}{9 x ^{2} - 1}

Solution steps

y^{'} (x) = \frac{x y ^{3}}{4}

Solve separable ODE:

y = - \frac{36}{9 x ^{2} - 1}

y = - \frac{36}{9 x ^{2} - 1}

\frac{s}{s ^{4} + 4}

\int \frac{2 x ^{4} + 4 x ^{3} - x}{x ^{3}} d x

\int 6^{x} e^{x} d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{17}{2}}} d x

y^{^{'}} + y = - 4 cos (x)

What is the solution for y^'=(xy^3)/4 ,y(0)=6 ?
The solution for y^'=(xy^3)/4 ,y(0)=6 is y=sqrt(-(36)/(9x^2-1))