1
$\begingroup$

I have an exercise and I have an idea of how to solve it, could you help me by giving me a hint?

$$(10-6y+e^{-3x})d x- 2 d y=0, \qquad y(0)=1$$

This is my attemp

$$ \frac{\partial F}{\partial x} = P(x,y)\\ \frac{\partial F}{\partial y} = Q(x,y)$$ Is not exact, so find $\mu(x)$ $$ \frac{\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}}{Q} = 3 $$ Therefore $$ \mu (x) = e^{\int 3 dx} \implies e^{3x}$$ Solving $$ (10-6y+e^{-3x})dx- 2 dy=0\\ e^{3x}[ (10-6y+e^{-3x})dx- 2 dy=0]\\ (10e^{3x}-6ye^{3x}+1)dx-2e^{3x}dy = 0\\ 10\int e^{3x}dx-6y\int e^{3x}dx+\int dx = 2e^{3x} \int ydy$$ And in the last line is my doubt, I dont know how to separate the equation.

Thank you

$\endgroup$
1
  • $\begingroup$ The second last line is an exact equation. I don't think it's separable. $\endgroup$
    – B. Goddard
    Nov 18, 2021 at 18:41

1 Answer 1

Reset to default
1
$\begingroup$

Sometimes those problems that are written as though they could be exact differentials are actually trick questions, and it's easier to solve by a different method.

So in your case, I would divide through by $dx$ and rewrite the equation as $$y' +3y =\frac{ e^{-3x}}{2} +5.$$ Is this enough of a hint for you to know what to do to solve?

$\endgroup$
1
  • 1
    $\begingroup$ Oh thank you, it had a trick and it was easier than I thought. $\endgroup$
    – Miguel JV
    Nov 18, 2021 at 19:49

You must log in to answer this question.

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