# Problem of the Week # 281 - Sep 18, 2017

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#### Ackbach

##### Indicium Physicus
Staff member
Here is this week's POTW (I will identify the problem source when I post the solution):

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Solve the ODE $(y^3+xy^2+y) \, dx + (x^3+x^2y+x) \, dy=0$.

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#### Ackbach

##### Indicium Physicus
Staff member
No one answered this week's POTW, which is Example 10.741 on page 87 in Tenenbaum and Pollard. The solution follows:

Comparing the ODE with $P(x,y) \, dx+Q(x,y) \, dy=0$, we see that $P(x,y)=y^3+xy^2+y$ and $Q(x,y)=x^3+x^2y+x$. Therefore,
$$\pd{P(x,y)}{y}=3y^2+2xy+1, \qquad \pd{Q(x,y)}{x}=3x^2+2xy+1,$$
which is not exact. We define
\begin{align*}
u&=xy \\
F(u)&=\frac{\pd{P(x,y)}{y}-\pd{Q(x,y)}{x}}{y Q(x,y)-x P(x,y)} \\
&=\frac{3y^2+2xy+1-3x^2-2xy-1}{x^3y+x^2y^2+xy-xy^3-x^2y^2-xy} \\
&=-\frac{3(x^2-y^2)}{xy(x^2-y^2)} \\
\end{align*}
Therefore, the integrating factor is $\displaystyle h(u)=e^{\int (-3/u) \, du}=e^{-3 \ln|u|}=u^{-3}=(xy)^{-3}.$ Multiplying the ODE through by this factor yields
$$\frac{1}{x^3y^3}(y^3+xy^2+y) \, dx+\frac{1}{x^3y^3}(x^3+x^2y+x) \, dy=0 \qquad \implies \qquad \left(x^{-3}+x^{-2}y^{-1}+x^{-3}y^{-2}\right) dx + \left(y^{-3}+x^{-1}y^{-2}+x^{-2}y^{-3}\right) dy=0.$$
Re-defining $P(x,y)=x^{-3}+x^{-2}y^{-1}+x^{-3}y^{-2}$ and $Q(x,y)=y^{-3}+x^{-1}y^{-2}+x^{-2}y^{-3}$ yields
$$\pd{P(x,y)}{y}=-x^{-2}y^{-2}-2x^{-3}y^{-3} , \qquad \pd{Q(x,y)}{x}=-x^{-2}y^{-2}-2x^{-3}y^{-3},$$
which is exact as required. To solve, we go through the usual steps:
$$f(x,y)=\int P(x,y) \, dx+R(y)=\frac{x^{-2}}{-2}+\frac{x^{-1}y^{-1}}{-1}+\frac{x^{-2}y^{-2}}{-2}+R(y)=-\frac12 x^{-2}-x^{-1}y^{-1}-\frac12 x^{-2}y^{-2}+R(y).$$
Differentiating w.r.t. $y$ yields
$$\pd{f(x,y)}{y}=x^{-1}y^{-2}+x^{-2}y^{-3}+R'(y)=Q(x,y)=y^{-3}+x^{-1}y^{-2}+x^{-2}y^{-3}.$$
By inspection, $R'(y)=y^{-3},$ implying that $R(y)=-\dfrac{y^{-2}}{2}.$ Hence, the solution is
$$-\frac12 x^{-2}-x^{-1}y^{-1}-\frac12 x^{-2}y^{-2}-\dfrac{y^{-2}}{2}=C, \qquad \text{or} \qquad \frac{1}{x^2}+\frac{2}{xy}+\frac{1}{x^2 y^2}+\frac{1}{y^2}=C,$$
where we have absorbed a $-2$ into the $C$.

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