What is the solution for (x^2+2y^2)=xyy^',y(-1)=1 ?

The solution for (x^2+2y^2)=xyy^',y(-1)=1 is y=-xsqrt(-1+2x^2)

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(x^2+2y^2)=xyy^',y(-1)=1

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

y = - x - 1 + 2 x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = - x - 1 + 2 x^{2}

y = - x - 1 + 2 x^{2}

\int z d z

\int_{3}^{\infty} \frac{4}{x ^{2} - x} d x

\int 8 x ln (2 x) d x

x \to 1 lim (x^{2} + x - 3)

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

What is the solution for (x^2+2y^2)=xyy^',y(-1)=1 ?
The solution for (x^2+2y^2)=xyy^',y(-1)=1 is y=-xsqrt(-1+2x^2)