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I am given this equation

$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$

where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $\mu=x^ny^m$.

I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $\mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!

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  • $\begingroup$ Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal. $\endgroup$ Oct 1 '17 at 7:51
  • $\begingroup$ After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$? $\endgroup$ Oct 1 '17 at 8:11
  • $\begingroup$ You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$. $\endgroup$
    – user436658
    Oct 1 '17 at 8:19
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In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12x\dfrac{1}{y})$ we have $$p(z)=\dfrac{M_y-N_x}{Ny-Mx}=\dfrac{\frac{12x}{y}}{4x^2}=\dfrac{3}{xy}=\dfrac{3}{z}$$ means your integrating factor is of the form $\mu(z)=\mu(xy)$. So $$I=e^{\int p(z)dz}=z^3=(xy)^3$$ and finally the answer is $$\color{blue}{x^4y^4-2x^5y^3=C}$$

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write your equation in the form $$y'(x)=-\frac{y(x)}{x}\frac{\left(-5+\frac{2y(x)}{x}\right)}{\left(-3+\frac{2y(x}{x}\right)}$$ and set $$u=\frac{y(x)}{x}$$

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The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form: $$\frac{dy}{dx}=g(\frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble. Then $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$ $$v+x\frac{dv}{dx}=g(v)$$ Which is separable. $$\frac{dv}{v-g(v)}+\frac{1}{x}dx=0$$

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The equation is of the form $M(x,y)dx + N(x,y)dy = 0$. Set $M(x,y) = 0$ to obtain $y_0(x) = \frac{5x}{2}$. Set $N(x,y) = 0$ to obtain $x_0(y) = \frac{2y}{3}$.

Let $g(y) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$, given $x = x_0(y)$. Similarly, let $f(x) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{-N}$, given $y = y_0(x)$. We obtain $g(y) = \frac{3}{y}$ and $f(x) = \frac{3}{x}$

Note that $f(x)$ and $g(y)$ satisfy $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = M(x,y)g(y) - N(x,y)f(x)$$. Hence an integrating factor, $\mu$, can be found by $e^{\int f(x)dx + \int g(y)dy + c}$. Doing this gives us $\mu = kx^3y^3$, for arbitrary k.

The method used here can be found at https://www.ijesm.co.in/current_issue.php

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