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

The solution for xdx+(x^2y+4y)dy=0,y(4)=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,y(4)=0

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

f (x) = 3 ln (x^{5}), at \frac{15}{e}

\int x^{2} + \frac{1}{x ^{2}} d x

\int (x + 1) sin (5 x) d x

\frac{d y}{d x} = \frac{cos ( 20 x )}{e ^{20 y}}, y (0) = 2

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