Integrating Factor: Solve x ln(x) dy/dx = xe^x

[tiaborrego]

How do you solve x ln(x) dy/dx = xe^x using the integrating factor?
So far, I have put it into standard form.
dy/dx + y/(xln(x))=xe^x/(x(ln(x))

 

[MarkFL]

I've moved this thread here to "Differential Equations" as it's a better fit for the problem.

I am assuming the original equation is:

\(\displaystyle x\ln(x)\d{y}{x}+y=xe^x\)

So that we then may write:

\(\displaystyle \d{y}{x}+\frac{1}{x\ln(x)}y=\frac{e^x}{\ln(x)}\)

Have we lost any trivial solutions by dividing through by $x\ln(x)$?

We compute the integrating factor as follows:

\(\displaystyle \mu(x)=\exp\left(\int\frac{1}{x\ln(x)}\,dx\right)\)

Can you find a $u$-substitution that will aid in this computation?

 

[tiaborrego]

Ok, so I think I found the u substitution. I think it is:
u= x* ln[x]
du = 1+ ln[x] dx
Therefore,
I(x)=e^ln[x*ln(x)] = x* ln[x]

Is this right?

 

[MarkFL]

No, what I would use is:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

and so we have:

\(\displaystyle \mu(x)=\exp\left(\frac{1}{u}\,du\right)\)

Now it should be easy to proceed. :)

 

[tiaborrego]

So, I tried the u substitution again, and this time I got:
(1/x)*e^{-1/x^2}
for I. Is this right?

My process:
u = ln(x)
du= 1/x dx
I= e^{1/u*du}
= e^{ln(u)*du}
=e^{ln[ln(x)]*1/x}
=(1/x)*e^{-1/x^2}

 

[MarkFL]

No, first let's observe that we must have $0<x$, from the original equation. When we divided through by $x\ln(x)$, we must note that we cannot have $x=0$ (already taken care of by the domain), and we cannot have $x=1$, so we are eliminating the trivial solution $y=e$. We may find we can include this back into the general solution.

Now, with that out of the way we note:

\(\displaystyle \mu(x)=\exp\left(\frac{1}{u}\,du\right)=e^{\ln(u)}=u=\ln(x)\)

Thus, our ODE becomes:

\(\displaystyle \ln(x)\d{y}{x}+\frac{1}{x}y=e^x\)

If we computed the integrating factor correctly, we should not find that the left side of the equation is:

\(\displaystyle \frac{d}{dx}\left(\mu(x)\cdot y\right)\)

Is this what we have?

 

[tiaborrego]

Ok, so I think I get it now.
So I solved for I
I= e^{x}
and thus, the general solution is:
y(x)=e^x/ln[x] + c/ln[x]

 

[MarkFL]

Ok, so I think I get it now.
So I solved for I
I= e^{x}
and thus, the general solution is:
y(x)=e^x/ln[x] + c/ln[x]

Yes, we have:

\(\displaystyle \frac{d}{dx}\left(\ln(x)y\right)=e^x\)

And when we integrate w.r.t. x, we obtain:

\(\displaystyle \ln(x)y=e^x+C\)

Hence:

\(\displaystyle y(x)=\frac{e^x+C}{\ln(x)}\)

But, we must also include the trivial solution $y\equiv e$ that was lost during the process. You should verify that this solution also satisfies the given ODE.