# Thread: How to solve this unexact differential equation?

1. ## How to solve this unexact differential equation?

1dx + (x/y-siny)dy = 0 ---> not exact

M(x,y) = 1 and N(x,y) = (x/y - siny)

δM/δy = 0 and δN/δx = (1/y)

To find the integrating factor;

g(x) = (1/N)*(δM/δy - δN/δx) and μ = exp[∫g(x)dx]

g(x) = 1/(x/y-siny)*(0-(1/y)

= 1/(x/y-siny)*(-1/y)

∫g(x)dx = ∫[-1/y / (x/y - siny)]dx = -ln(x-ysiny)

μ= e^-[ln(x-ysiny)]= 1/(x-ysiny)

but I couldn't figure out the rest of the question... Is there an error or someting? I'm not sure...Can anyone help me to solve this question?

2. I found the integrating factor μ= e^-[ln(x-ysiny)]= 1/(x-ysiny) but after that,

δM/δy is still not equal to δN/δx when I multiplied with μ...

3. Hi. I get $\mu=y$ as the integrating factor. Also, try and learn Latex cus' it helps if your math is beautiful like:

$\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=-\frac{1}{y}\rightarrow g(y)=-\frac{1}{y}\therefore \mu=\text{exp}\left(\int \frac{1}{y}dy\right)=y$

4. Originally Posted by shawsend
Hi. I get $\mu=y$ as the integrating factor. Also, try and learn Latex cus' it helps if your math is beautiful like:

$\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=-\frac{1}{y}\rightarrow g(y)=-\frac{1}{y}\therefore \mu=\text{exp}\left(\int \frac{1}{y}dy\right)=y$
I've probably made an error. The question was in fact very easy, how couldn't I think of this easier method...Thank you for your assistane

$\frac{dx}{dy} + \frac{x}{y} = \sin y$
it's linear in $x$.