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[SOLVED] Initial Value Problem

alane1994

Active member
Oct 16, 2012
126
Ok, I have a practice exam... My professor gave out a copy with worked out examples. There is one where I don't get his logic at all. I was wondering if you guys could explain it to me?


\(\displaystyle (\frac{1}{t}+2y^2t)dt+(2yt^2-\cos(y))dy=0\)

First, he put \(\text{Assume t>0}\)
???

\(\displaystyle \text{Since} \frac{\partial}{\partial y}(\frac{1}{t}+2y^2t)=4yt=\frac{\partial}{\partial t}(2yt^2-\cos(y))\)

\(\displaystyle F(t,y)=\int (\frac{1}{t}+2y^2t)dt=\ln(t)+y^2t^2+f(y)\)
\(\displaystyle ~~~~~~~~~~~=\int (2yt^2-\cos(y))dy=y^2ts-\sin(y)+g(t)\)

From those (not sure how to do two lines into right brace).

\(\ln(t)+t^2y^2-\sin(y)=C\)
\(y(1)=\pi\)
\(0+\pi^2-0=C\)

\(\therefore \ln(t)+t^2+y^2-\sin(y)=\pi^2\)
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Re: Ivp

Ok, I have a practice exam... My professor gave out a copy with worked out examples. There is one where I don't get his logic at all. I was wondering if you guys could explain it to me?


\(\displaystyle (\frac{1}{t}+2y^2t)dt+(2yt^2-\cos(y))dy=0\)

First, he put \(\text{Assume t>0}\)
???

\(\displaystyle \text{Since} \frac{\partial}{\partial y}(\frac{1}{t}+2y^2t)=4yt=\frac{\partial}{\partial t}(2yt^2-\cos(y))\)

\(\displaystyle F(t,y)=\int (\frac{1}{t}+2y^2t)dt=\ln(t)+y^2t^2+f(y)\)
\(\displaystyle ~~~~~~~~~~~=\int (2yt^2-\cos(y))dy=y^2ts-\sin(y)+g(t)\)

From those (not sure how to do two lines into left brace).

\(\ln(t)+t^2y^2-\sin(y)=C\)
\(y(1)=\pi\)
\(0+\pi^2-0=C\)

\(\therefore \ln(t)+t^2+y^2-\sin(y)=\pi^2\)
An ODE written in the form...


$\displaystyle A (t,y)\ dt + B(t,y)\ dy = 0\ (1)$

... where $\displaystyle \frac{\partial A}{\partial y}= \frac{\partial B}{\partial t}$ is called 'exact differential' and its solution, with initial condition $y(t_{0})= y_{0}$ is...


$\displaystyle \int_{t_{0}}^{t} A(u,y)\ du + \int_{y_{0}}^{y} B(t_{0},v)\ dv = c\ (2)$

In Your case is $\displaystyle A(t,y) = \frac{1}{t} + 2\ y^{2}\ t$ and $\displaystyle B (t,y) = 2\ y\ t^{2} - \cos y$ so that $\displaystyle \frac{\partial A}{\partial y} = \frac{\partial B}{\partial t}$ and You have an exact differential. The general solution is given from (2)...

$\displaystyle \int_{t_{0}}^{t} (\frac{1}{u} + 2\ u\ y^{2})\ du + \int_{y_{0}}^{y} (2\ t_{0}^{2}\ v - \cos v)\ dv= c\ (3)$

... and the integration of (3) is left to You...

Kind regards

$\chi$ $\sigma$
 

alane1994

Active member
Oct 16, 2012
126
Where did the the

\(y^2ts\)

come from?

How did terms cancel like that? It seems like they shouldn't.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: Ivp

This is a pretty standard "exact differential" problem. First, the "Assume t> 0" is to avoid t= 0 which, because of the "1/t", would cause trouble. Of course there would be another solution if we were to "assume t< 0".

(I just noticed that I have used "x, y" where you have "t, y". Assume my "x" is your "t".)

If F(x, y) is a differentiable function of x and y, then its differential is [tex]\frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]. As long as those partial derivatives are smooth, we must have the "mixed derivatives" the same: [tex]\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial^2 F}{\partial y\partial x}[/tex] (where the order of "[tex]\partial x[/tex]" and "tex]\partial y[/tex]" indicates the order of differentiation).

Now, we can write "u(x,y)dx+ v(x,y)dy" which looks like a differential- but it may not be! (We would say it is not "exact".) If this were really a differential, We would have to have [tex]\frac{\partial F}{\partial x}=u[/tex] and [tex]\frac{\partial F}{\partial y}= v[/tex] then [tex]\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial u}{\partial y}= \frac{\partial v}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}[/tex].

That is what is called the "mixed partials test" and they are showing after "Since".

Because that is true, we know there exist some function, F(x,y) such that [tex]\frac{\partial F}{\partial x}= \frac{1}{x}+ 2y^2x[/tex]. Integrating (with respect to x, treating y as a constant) [tex]F= ln(x)+ y^2x^2+ \phi(y)[/tex]. (Since we are treating y as a constant, the "constant of integration" may depend on y. That is the "[tex]\phi(y)[/tex]" above.)

Now, differentiate [tex]F= ln(x)+ y^2x^2+ \phi(y)[/tex] with respect to y: [tex]\frac{\partial F}{\partial y}= 2y^2x+ \frac{d\phi}{dy}[/tex]. (Notice that the derivative of [tex]\phi[/tex] is an ordinary derivative, not a partial derivative because [tex]\phi[/tex] is a function of the single variable y.)

But we know, from the equation, [tex]\frac{\partial F}{\partial y}= 2yx^2- cos(y)[/tex] so we must have [tex]2y^2x+ \frac{d\phi}{dy}= 2yx^2- cos(y)[/tex]. Notice that the only terms involving x, [tex]2yx^2[/tex], cancel! That had to happen because of the "mixed derivatives" condition. That leaves just [tex]\frac{d\phi}{dy}= -cos(y)[/tex] so that [tex]\phi(y)= -sin(y)+ C[/tex] (C really is a constant here) so that
[tex]F(x,y)= ln(x)+ y^2x^2- sin(y)+ C[/tex].

Since the differential equation was "dF= 0", F must be a constant: [tex]F(x,y)= ln(x)+ y^2x^2- sin(y)+C= c[/tex] or just [tex]F(x, y)= ln(x)+ y^2x^2- sin(y)= C'[/tex] where I have combined the two constants: C'= c- C.

Dropping the "F", which was not part of the original equation, [tex]ln(x)+ y^2x^2- sin(y)= C[/tex].

(Replacing x with your t:
[tex]ln(t)+ y^2t^2- sin(y)= C[/tex].)

(I suspect the "[tex]y^2ts[/tex]" was a typo. It should be just "[tex]y^2t[/tex]".)
 
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alane1994

Active member
Oct 16, 2012
126
Thank you guys very much! Makes a lot more sense now!