Answer to Question #250467 in Differential Equations for Jia

Question

exact (x^2+y^2-5)dx=(y+xy)dy y(0)=1

Answer 1

$$(x^2+y^2-5)dx+(-y-xy)dy=0$$

$$M(x, y)=x^2+y^2-5, M_y=2y$$

$$N(x, y)=-y-xy, N_x=-y$$

$$M_y\not=N_x$$

$$\dfrac{M_y-N_x}{N}=\dfrac{2y-(-y)}{-y-xy}=-\dfrac{3}{x+1}=P(x)$$

Integrating factor

$$\mu(x)=e^{\int(-{3 \over x+1})dx}=\dfrac{1}{(x+1)^3}$$

$$\dfrac{x^2+y^2-5}{(x+1)^3}dx+\dfrac{-y-xy}{(x+1)^3}dy=0$$

$$M(x, y)=\dfrac{x^2+y^2-5}{(x+1)^3}, M_y=\dfrac{2y}{(x+1)^3}$$

$$N(x, y)=-\dfrac{y}{(x+1)^2}, N_x=\dfrac{2y}{(x+1)^3}$$

$$M_y=\dfrac{2y}{(x+1)^3}=N_x$$

$$f(x, y)=\int(-\dfrac{y}{(x+1)^2})dy+g(x)$$

$$=-\dfrac{y^2}{2(x+1)^2}+g(x)$$

$$f_x=\dfrac{y^2}{(x+1)^3}+g'(x)=\dfrac{x^2+y^2-5}{(x+1)^3}$$

$$g'(x)=\dfrac{x^2-5}{(x+1)^3}$$

$$g(x)=\int\dfrac{x^2-5}{(x+1)^3}dx=\int\dfrac{x^2+2x+1-2x-2-4}{(x+1)^3}dx$$

$$=\int\dfrac{1}{x+1}dx-2\int\dfrac{1}{(x+1)^2}dx-4\int\dfrac{1}{(x+1)^3}dx$$

$$=\ln(|x+1|)+\dfrac{2}{x+1}+\dfrac{2}{(x+1)^2}-C$$

$$-\dfrac{y^2}{2(x+1)^2}+\ln(|x+1|)+\dfrac{2}{x+1}+\dfrac{2}{(x+1)^2}=C$$

The initial condition $y(0)=1$

$$-\dfrac{(1)^2}{2(0+1)^2}+\ln(|0+1|)+\dfrac{2}{0+1}+\dfrac{2}{(0+1)^2}=C$$

$$C=\dfrac{7}{2}$$

The solution of the initial value problem

$$-\dfrac{y^2}{2(x+1)^2}+\ln(|x+1|)+\dfrac{2}{x+1}+\dfrac{2}{(x+1)^2}=\dfrac{7}{2}$$