Question

{F} Y(4x+y)dx-2(x^2-y)dy=0

Answer 1

Solution:

$\left(2 y-2 x^{2}\right) d y+\left(y^{2}+4 x y\right) d x=0$

Equation in total differentials $\mu(x, y) M(x, y) \mathrm{d} y+\mu(x, y) N(x, y) \mathrm{d} x=0$  

where $M(x, y)=2 y-2 x^{2}\ and\ N(x, y)=y^{2}+4 x y$

Check for full differential: $M(x, y)_{x}^{\prime}=-4 x \neq 2 y+4 x=N(x, y)_{y}^{\prime}$  

Search for an integrating factor $\mu(x, y)^{?} M(x, y)_{x}^{\prime}=\frac{\partial M}{\partial x}\ and\ N(x, y)_{y}^{\prime}=\frac{\partial N}{\partial y}$  From condition: $M \frac{\partial \mu}{\partial x}-N \frac{\partial \mu}{\partial y}=\mu\left(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial x}\right)$

Let $\mu(x, y)=\mu(y) \Rightarrow \frac{\partial \mu}{\partial x}=0$  then the condition becomes: $\frac{1}{\mu} \frac{\mathrm{d} \mu}{\mathrm{d} y}=-\frac{1}{N}\left(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial x}\right)$  where right side - function from y

$\begin{gathered} \int \frac{\mathrm{d} \mu}{\mu}=\int-\frac{1}{N}\left(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial x}\right) \mathrm{d} y=\int-\frac{2}{y} \mathrm{~d} y=-2 \ln (y) \\ \ln (\mu)=-2 \ln (y) \Rightarrow \mu=\frac{1}{y^{2}} \end{gathered}$

Multiply the differential equation by $\frac{1}{y^{2}}$

$\left(\frac{2}{y}-\frac{2 x^{2}}{y^{2}}\right) \mathrm{d} y+\left(\frac{4 x}{y}+1\right) \mathrm{d} x=0, \quad$  lost when dividing solution: $\frac{1}{y^{2}} \rightarrow y=0$

Equation in total differentials $M(x, y) \mathrm{d} y+N(x, y) \mathrm{d} x=0$  

where $M(x, y)=\frac{2}{y}-\frac{2 x^{2}}{y^{2}}\ and\ N(x, y)=\frac{4 x}{y}+1$

Check for full differential: $M(x, y)_{x}^{\prime}=N(x, y)_{y}^{\prime}=-\frac{4 x}{y^{2}}$

Find $F(x, y): \mathrm{d} F(x, y)={F}_{y}^{\prime} \mathrm{d} y+F_{x}^{\prime} \mathrm{d} x$

$\begin{gathered} F(x, y)=\int N(x, y) \mathrm{d} x=\int_{y}^{\prime} \frac{4 x}{y}+1 \mathrm{~d} x=\frac{2 x^{2}}{y}+x+C_{y} \\ \left(\frac{2 x^{2}}{y}+x\right)_{y }^{\prime}=-\frac{2 x^{2}}{y^{2}} \\ C_{y}=\int M(x, y)-\left(\frac{2 x^{2}}{y}+x\right)_{y}^{\prime} \mathrm{d} y=\int \frac{2}{y} \mathrm{~d} y=2 \ln (y) \\ F(x, y)=\frac{2 x^{2}}{y}+x+C_{y}=2 \ln (y)+\frac{2 x^{2}}{y}+x \end{gathered}$

Thus, the solution is $2 \ln (y)+\frac{2 x^{2}}{y}+x=C$