Answer to Question #277442 in Differential Equations for Jane

Question

Find the solution of Bernoulli Equation

Given:

2xdy+y(y^2lnx-1)dx=0

Answer 1

$2xdy+y(y^2lnx-1)dx=0\\ 2xdy+(y^3lnx-y)dx=0$

Multiply by $\frac{1}{y^3}$ both sides, we get

$\frac{2xdy}{y^3}+(lnx-\frac{1}{y^2})dx=0$

$\begin{aligned} &\text { Equation in total differentials } M(x, y) \mathrm{d} y+N(x, y) \mathrm{d} x=0 \\ &\text { where } M(x, y)=\frac{2 x}{y^{3}} \text { and } N(x, y)=\ln (x)-\frac{1}{y^{2}} \\ &\text { Check for full differential: } M(x, y)_{x}^{\prime}=N(x, y)_{y}^{\prime}=\frac{2}{y^{3}} \\ &\text { Find } F(x, y): \mathrm{d} F(x, y)=F_{y}^{\prime} \mathrm{d} y+F_{x}^{\prime} \mathrm{d} x \\ &F(x, y)=\int N(x, y) \mathrm{d} x=\int \ln (x)-\frac{1}{y^{2}} \mathrm{~d} x=x \ln (x)-\frac{x}{y^{2}}-x+C_{y} \\ &\qquad\left(x \ln (x)-\frac{x}{y^{2}}-x\right)_{y}^{\prime}=\frac{2 x}{y^{3}} \\ &\qquad \begin{array}{c} C_{y}=\int M(x, y)-\left(x \ln (x)-\frac{x}{y^{2}}-x\right)_{y}^{\prime} \mathrm{d} y=\int 0 \mathrm{~d} y=0 \\ F(x,y)= x \ln (x)-\frac{x}{y^{2}}-x+C_{y}=x \ln (x)-\frac{x}{y^{2}}-x \end{array} \end{aligned}$

$x \ln (x)-\frac{x}{y^{2}}-x=C$