What is the solution for xdx+(x^2y+4y)dy=0 ?

The solution for xdx+(x^2y+4y)dy=0 is y=sqrt(-ln(x^2+4)+c_{1)},y=-sqrt(-ln(x^2+4)+c_{1)}

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xdx+(x^2y+4y)dy=0

x d x + (x^{2} y + 4 y) d y = 0

y = - ln (x^{2} + 4) + c_{1}, y = - - ln (x^{2} + 4) + c_{1}

Solution steps

x d x + (x^{2} y + 4 y) d y = 0

Solve separable ODE:

y = - ln (x^{2} + 4) + c_{1}, y = - - ln (x^{2} + 4) + c_{1}

y = - ln (x^{2} + 4) + c_{1}, y = - - ln (x^{2} + 4) + c_{1}

y^{^{'}} + 5 y = 2

\int \frac{( x - 1 ) ^{3}}{x} d x

\frac{d}{d x} (A (e^{2 x} + 2 e^{2 x} x))

\frac{d}{d x} (- 2 cot (x) + 6 x^{4} csc (x))

f (x) = x^{\frac{1}{9}}

What is the solution for xdx+(x^2y+4y)dy=0 ?
The solution for xdx+(x^2y+4y)dy=0 is y=sqrt(-ln(x^2+4)+c_{1)},y=-sqrt(-ln(x^2+4)+c_{1)}