# Finding explicit maximal solution of an IVP using Exact Diff Eqs

#### nathancurtis11

##### New member
I need to find the explicit maximal solution of an IVP using exact Diff Eqs:

The IVP is given as:

{xexyy'-cos(x)+yexy=0
{y(0)=1

So I know at first I need to get the implicit solution by getting that:
A(x,y) = xexy
B(x,y) = -cos(x)+yexy

I know I need to find the partial derivative of A(x,y) with respect to x then the partial derivative of B(x,y) with respect to y.

After this I can't quite figure out how to get this one solved implicitly first, but the real problem lies with getting it explicitly defined (hence giving us the explicit maximal solution) and the maximal open interval after that.

Last edited:

#### MarkFL

Staff member
I would first write the ODE in differential form:

$$\displaystyle \left(ye^{xy}-\cos(x) \right)dx+\left(xe^{xy} \right)dy=0$$

Next we need to compute:

$$\displaystyle \frac{\partial}{\partial y}\left(ye^{xy}-\cos(x) \right)=e^{xy}(xy+1)$$

$$\displaystyle \frac{\partial}{\partial x}\left(xe^{xy} \right)=e^{xy}(xy+1)$$

Thus, we see that the ODE is exact. And so we must have:

$$\displaystyle \frac{\partial F}{\partial x}=ye^{xy}-\cos(x)$$

Integrate this with respect to $x$ to get:

(1) $$\displaystyle F(x,y)=\int ye^{xy}-\cos(x)\,dx+g(y)$$

To determine $g(y)$, take the partial derivative with respect to $y$ of both sides and substitute $$\displaystyle xe^{xy}$$ for $$\displaystyle \frac{\partial F}{\partial y}$$. Then solve for $g'(y)$ to obtain $g(y)$ up to a numerical constant. Then substitute for $g(y)$ in (1) to obtain $F(x,y)$. The solution to the ODE is then given implicitly by:

$$\displaystyle F(x,y)=C$$

So, what do you find?

#### nathancurtis11

##### New member
Okay so I implicitly solved it and got:

exy-sin(x)=1

Then I was able to pretty easily solve it explicitly from there:

$$\displaystyle y=(ln|1+sin(x)|)/x$$

Now after this I need to find the maximal open interval and thats where I'm really really stuck.

#### nathancurtis11

##### New member
Looking at my notes, 0 is supposed to be included because its a solution to begin the problem, yet 0 is undefined in this specific case if the above explicit solution is correct. So, I'm a bit confused even more so now.

#### MarkFL

Staff member
Perhaps this is overcome by:

$$\displaystyle \lim_{x\to0}\frac{\ln(1+\sin(x))}{x}=1$$

I am unfamiliar with the term "maximal explicit solution" so I would wait until someone who is familiar with it can weigh in. #### nathancurtis11

##### New member
So the maximal solution was basically just asking on what open interval is the explicit solution maximized (so basically the maximum domain on which it is defined on) on including the original initial condition. What really confused me though is the fact that there is an infinite amount of numbers in which (sin x) was not equal to -1, therefore defined, and also an infinite amount of numbers in which (sin x) was equal to -1, therefore undefined. So I ended up just giving the solution (-inf, inf) where sinx is not equal to -1 and got full credit. #### Ackbach

##### Indicium Physicus
Staff member
So I ended up just giving the solution (-inf, inf) where sinx is not equal to -1 and got full credit. Just a small note here: it is MHB policy not knowingly to help with any problem that counts towards a student's grade. Best thing to do is ask for help on a similar problem to one on which you are stuck.