0
$\begingroup$

The following problem is from a text book on differential equations but it is not an homework problem.

Problem: Solve the following differential equation by first finding an integrating factor: \begin{equation*} (5xy + 4y^2 + 1) dx + ( x^2 + 2xy) dy = 0 \end{equation*}

I have no idea on how to find an integrating factor. I am hoping somebody here can help me. I thank the group in advance for their responses.

Bob

$\endgroup$

2 Answers 2

1
$\begingroup$

We guess that the integrating factor will be $I(x)$, a function of $x$ only. After multiplying by $I(x)$, the equation will be exact if $$\frac{\partial}{\partial y}\bigl((5xy+4y^2+1)I(x)\bigr)=\frac{\partial}{\partial x}\bigl((x^2+2xy)I(x)\bigr)\ .$$ If you do the differentiations and simplify, you should end up with $$\frac{dI}{dx}=\frac{3I}{x}\ .$$ You can solve this to find an integrating factor.

$\endgroup$
3
  • $\begingroup$ David, I worked out your differential equation and found that: I=e^(cx^3) which is not a valid integrating factor. I am thinking that the integrating factor is not a function of x alone. $\endgroup$
    – Bob
    Jun 18, 2014 at 2:00
  • $\begingroup$ I think you need to check your solution for $I$, you should have got $x^3$ not $e^{x^3}$. $\endgroup$
    – David
    Jun 18, 2014 at 2:21
  • $\begingroup$ You are right, it is $x^3$. $\endgroup$
    – Bob
    Jun 18, 2014 at 23:24
0
$\begingroup$

If this equation is not exact, you can find an integrate factor by

$\mu = \dfrac{\frac{{\partial}M}{{\partial}y} - \frac{{\partial}N}{{\partial}x}}{M}$

$\endgroup$
1
  • $\begingroup$ My text book says that an integrating factor can be found as suggested by user115595 but it says to use that method only if u depends upon x only. However, in this case, u is going to depend on both and x and y. Therefore, it is my understanding that the solution suggested by user115595 is wrong. Is there something I am missing? $\endgroup$
    – Bob
    Jun 18, 2014 at 1:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .