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Solve the equation $\frac{d y}{d x}=\frac{e^{y^{2}}}{y-2 x y e^{y^{2}}}$

$M d x+N d y=0$

$\frac{d y}{d x}=\frac{e^{y^{2}}}{y-2 x y e^{y^{2}}}$

$e^{y^{2}} d x=\left(y-2 x y e^{y^{2}}\right) d y$

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

(2) $\frac{d M}{d y}=\frac{d N}{d x}$

$\frac{d M}{d y}=2 y e^{y^{2}}$

$\frac{d N}{d x}=-\left(0-2 y e^{y^{2}}\right)=2 y e^{y^{2}}$

Exact diffrential equation

(1) $\psi(x, y)=\int M \cdot d x+g(y)$

$=\int e^{y^{2}} d x+g(y)=e^{y^{2}} x+g(y)$

$\frac{d \psi}{d y}=2 x y e^{y^{2}}+g^{\prime}(y) \rightarrow=N$

$2 x y e^{y^{2}}+g^{\prime}(y)=N=-\left(y-2 x y e^{y^2}\right)$

$2 x y e^{y^{2}}+g^{\prime}(y)=-y+2 x y e^{y^2}$

$g(y)=\int-y d y=-\frac{y^{2}}{2}$

$\psi(x, y)=x e^{y^{2}}-\frac{1}{2} y^{2}=C$

Find the value of $(b)$ for which the DE is exact then solve it
$\left(y e^{2 x y}+3 x\right)+\left(b x e^{2 x y}\right) y^{\prime}=0$

(1) $\left(y e^{2 x y}+3 x\right)+\left(b x e^{2 x y}\right) y^{\prime}=0$

(2) $\frac{d M}{d y}=\frac{d N}{d x}$

$\frac{d M}{d y}=(1) e^{2 x y}+(2 x) y e^{2 x y}$

$\frac{d N}{d x}=b e^{2 x y}+b(2 y) x e^{2 x y}$

Exact

$b=1$

(1) $\psi(x, y)=\int\left(y e^{2 x y}+3 x\right) d x+g(y)$

$\psi(x, y)=\int {\frac{2}{2}}\left(y e^{2 x y}+3 x\right) d x+g(y)$

$=\frac{1}{2} e^{2 x y}+\frac{3 x^{2}}{2}+g(y)$

$\frac{d \psi}{d y}=\frac{1}{2} e^{2 x y}(2 x)+0+g^{\prime}(y)$

$x e^{2 x y}+g^{\prime}(y)=x e^{2 x y}$

$g^{\prime}(y)=0$

$\int g^{\prime}(y)=\int 0 d y$

$g(y)=c$

$\psi(x, y)=\frac{1}{2} e^{2 x y}+\frac{3}{2} x^{2}=c$