What is the solution for dx+ydy=x^2ydy ?

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

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dx+ydy=x^2ydy

d x + y d y = x^{2} y d y

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

Solution steps

d x + y d y = x^{2} y d y

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{x}{y}, y = 2, x = 1

3 y^{2} \frac{d y}{d x} = (5 x)

y^{^{'}} = \frac{- 2 y + x y}{x ^{2}}

\frac{d y}{d x} = \frac{( y + 3 ) ( x - 1 )}{( y - 2 ) ( x + 4 )}

\frac{d y}{d x} y = \frac{5}{cos ( x )} + \frac{1}{tan ( x )}

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