Finding the explicit solution to the IVP

[shamieh]

Find the explicit solution to the IVP.

$xdx + ye^{-x}dy=0$, $y(0) =1$
so I did some manipulation to get
$ye^{-x}dy= -xdx$ ==> $\frac{dy}{dx}=\frac{-x}{ye^{-x}}$

but now I'm confused on what to do. What I found above is the implicit solution right? So do I just need to get $y'$ on the left side by multiplying through with a $dx$ and then just plug a $0$ in for $x$ and a $1$ in for $y$ to get the explicit solution??

 

[MarkFL]

An implicit solution to an ODE is a relationship derived from the ODE in which you cannot solve for either variable, whereas an explicit solution is one in which you can solve for one of the variables.

I think what I would do is separate the variables to obtain:

$$y\,dy=-xe^x\,dx$$

Now integrate both sides:

$$\int_1^y u\,du=\int_x^0 ve^v\,dv$$

What do you find?

 

[shamieh]

Okay that's what I suspected. I got $c=1$ thus I got for my final explicit solution $y= \sqrt{2e^x-2xe^x-1}$

 

[MarkFL]

Okay that's what I suspected. I got $c=1$ thus I got for my final explicit solution $y= \sqrt{2e^x-2xe^x-1}$

I get the same. (Yes)