# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Di Qiu on October 04, 2019, 02:06:04 PM

Title: TUT0402 Quiz2
Post by: Di Qiu on October 04, 2019, 02:06:04 PM
Question: $$\frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} + \frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \cdot \frac{dy}{dx} = 0$$
First, we need to find $$M_y(x,y)$$
Where, \begin{align*} M_y(x,y) &= \frac{\partial}{\partial y}M(x,y) \\ &= \frac{\partial}{\partial y}\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}} \\ &= x \cdot \frac{\partial}{\partial y}(x^2+y^2)^{-\frac{3}{2}} \\ &= x \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2y \\ &= -\frac{3xy}{(x^2+y^2)^{\frac{5}{2}}} \end{align*}
Then for the $$N_x(x,y)$$
Where, \begin{align*} N_x(x,y) &= \frac{\partial}{\partial x} \\ &= \frac{\partial}{\partial x}\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \\ &= y \cdot \frac{\partial}{\partial x}(x^2+y^2)^{-\frac{3}{2}} \\ &= y \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2x \\ &= -\frac{3yx}{(x^2+y^2)^{\frac{5}{2}}} \end{align*}
Since $$M_y(x,y) = N_x(x,y)$$ are exact, there exist a Φ, where Φx = M, Φy = N.
Therefore, \begin{align*} \phi &= \int \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} dx \\ &=\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} dx + g(y) \end{align*}
$$\phi_y = \frac{y}{{(x^2+y^2)}^{\frac{2}{3}}} + g'(y)$$
Compare it to N(x,y), we know that g'(y) = 0, so g(y) = constant.
Therefore, $$\phi = \frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} + C$$
General Solution: $$\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} = C$$