MAT244--2019F > Quiz-2

TUT0401 quiz2 solution

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EroSkulled:
Solve$\frac{x}{(x^2+y^2)^\frac{3}{2}} + \frac{y}{(x^2+y^2)^\frac{3}{2}}\frac{dy}{dx}=0$

Let $M=\frac{x}{(x^2+y^2)^\frac{3}{2}}$ and $N=\frac{y}{(x^2+y^2)^\frac{3}{2}}$
We want to find $N_x$ and $M_y$

M_y=\frac{d}{dy}\frac{x}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}

N_x=\frac{d}{dx}\frac{y}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}

N_x=M_y

We conclude the given equation is exact. We integrate to find $\varphi(x,y)$
Note $\varphi_x(x,y)=M$, integrate both side, we get:

\varphi(x,y)=\int{Mdx}=\int{\frac{x}{(x^2+y^2)^\frac{3}{2}}dx}

Using u subsitution with $u=x^2+y^2$, we get:

\varphi(x,y)=\frac{1}{2}\int{\frac{1}{u^\frac{3}{2}}du}=-u^{-\frac{1}{2}}=-(x^2+y^2)^{-\frac{1}{2}}+h(y)

Note $\varphi_y(x,y)\equiv N$, then:

\varphi_y(x,y)=\frac{d}{dy}(-(x^2+y^2)^{-\frac{1}{2}}+h(y))

\varphi_y(x,y)=\frac{y}{(x^2+y^2)^\frac{3}{2}}+h'(y)\equiv \frac{y}{(x^2+y^2)^\frac{3}{2}}

Hence we can conlude $h'(y)=0$
Then $h(y)=C$

\varphi(x,y)=-(x^2+y^2)^{-\frac{1}{2}}+C

The implicit solutoin is $-(x^2+y^2)^{-\frac{1}{2}}=C$