MHB Finding the explicit solution to the IVP

  • Thread starter Thread starter shamieh
  • Start date Start date
  • Tags Tags
    Explicit Ivp
shamieh
Messages
538
Reaction score
0
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??
 
Physics news on Phys.org
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?
 
Okay that's what I suspected. I got $c=1$ thus I got for my final explicit solution $y= \sqrt{2e^x-2xe^x-1}$
 
shamieh said:
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)
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top