What is the solution for (x^2-5)yy^'=2x ?

The solution for (x^2-5)yy^'=2x is y=sqrt(2ln(x^2-5)+c_{1)},y=-sqrt(2ln(x^2-5)+c_{1)}

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(x^2-5)yy^'=2x

(x^{2} - 5) y y^{^{'}} = 2 x

y = 2 ln (x^{2} - 5) + c_{1}, y = - 2 ln (x^{2} - 5) + c_{1}

Solution steps

(x^{2} - 5) y y^{'} (x) = 2 x

Solve separable ODE:

y = 2 ln (x^{2} - 5) + c_{1}, y = - 2 ln (x^{2} - 5) + c_{1}

y = 2 ln (x^{2} - 5) + c_{1}, y = - 2 ln (x^{2} - 5) + c_{1}

\int \frac{x}{3} \cdot sin (x^{2}) d x

\int (x) d x

12 x^{2} - 8 x - 4

\int \frac{1}{e ^{x} - 1} d x

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 x)

What is the solution for (x^2-5)yy^'=2x ?
The solution for (x^2-5)yy^'=2x is y=sqrt(2ln(x^2-5)+c_{1)},y=-sqrt(2ln(x^2-5)+c_{1)}