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

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

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

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

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

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

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

$$g(y)=c$$

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

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