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How do you rewrite this equation in standard form?

find_the_fun

Active member
Feb 1, 2012
166
\(\displaystyle y dx-4(x+y^6)dy=0\)

so is it legal to add \(\displaystyle 4(x+y^6)dy\) to both sides of the equation? I thought it's bad to add integrands (i.e. dx or dy).

Or is it legal to divide both sides of the equation by dx? Would this result in
\(\displaystyle y-4(x+y^6)\frac{dy}{dx}=0\) or \(\displaystyle y-\frac{4(x+y^6)}{dx}\frac{dy}{dx}=0\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The given ODE is already written in differential form. To solve it, you need to determine if it is exact, and if not, to compute an integrating factor to make it exact.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
\(\displaystyle y dx-4(x+y^6)dy=0\)

so is it legal to add \(\displaystyle 4(x+y^6)dy\) to both sides of the equation? I thought it's bad to add integrands (i.e. dx or dy).

Or is it legal to divide both sides of the equation by dx? Would this result in
\(\displaystyle y-4(x+y^6)\frac{dy}{dx}=0\) or \(\displaystyle y-\frac{4(x+y^6)}{dx}\frac{dy}{dx}=0\)
I don't think that there is really a "standard" form to these. But typically they are written in both formats.

One problem: You have too many "dx"s in that last equation. The first form on that line is correct.

-Dan
 

find_the_fun

Active member
Feb 1, 2012
166
I'm still stuck. I'm getting the wrong answer.

\(\displaystyle ydx-4(x+y^6)dy=0\)
\(\displaystyle y-4(x+y^6)\frac{dy}{dx}=0\)
\(\displaystyle \frac{dy}{dx}-\frac{y}{4(x+y^6)}=0\) (1)
therefore \(\displaystyle P(x)=\frac{-1}{4(x+y^6)}\)

let \(\displaystyle \mu = e^{\int P(x) dx} = e^{\frac{-1}{4(x+y^6)}}dx = e^{\ln{|(x+y^6)^{\frac{1}{4}}|}}+C\) choose \(\displaystyle C=0\)

multiply both sides of (1) by integrating factor
\(\displaystyle \frac{dy}{dx} \frac{1}{(x+y^6)^{\frac{1}{4}}}-\frac{y}{4(x+y^6)} \frac{1}{(x+y^6)^{\frac{1}{4}}}=0\)
\(\displaystyle \frac{d}{dx}[\mu(x)y]=\frac{d}{dx}(x+y^6)^{\frac{-1}{4}}y=0\)
\(\displaystyle \int \frac{d}{dx}[(x+y^6)^{\frac{-1}{4}}y]dx = \int 0 dx\)
\(\displaystyle \frac{(x+y^6)^{-3}}{-3}+C=\frac{-1}{3(x+y^6)^3}+C\)
from this equation I can solve for y and get \(\displaystyle y=\sqrt[3]{\frac{-1}{3C}-x}\) and solving for x gives \(\displaystyle x=\frac{-1-3Cy^3}{3C}\)

The back of book gives
\(\displaystyle x=2y^6+cy^4\)
First of all I don't get that answer. Second, why do they solve for x and not y?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You cannot treat this equation as linear, since we cannot write in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

We can however, use an integrating factor to get an equivalent exact equation.

We are given:

\(\displaystyle y\,dx-4\left(x+y^6 \right)\,dy=0\)

We see that:

\(\displaystyle \frac{\partial}{\partial y}(y)=1\)

\(\displaystyle \frac{\partial}{\partial x}\left(-4\left(x+y^6 \right) \right)=-4\)

Thus, the equation is inexact. So we consider:

\(\displaystyle \frac{\frac{\partial}{\partial x}\left(-4\left(x+y^6 \right) \right)-\frac{\partial}{\partial y}(y)}{y}=-\frac{5}{y}\)

So our integrating factor is:

\(\displaystyle \mu(y)=e^{\int -\frac{5}{y}\,dy}=y^{-5}\)

And thus our ODE becomes (observing that we are eliminating the trivial solution $y\equiv0$):

\(\displaystyle y^{-4}\,dx-4\left(xy^{-5}+y \right)\,dy=0\)

And we can easily see that this equation is exact. We know that the solution must therefore be given implicitly by:

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

And because the equation is exact, we know:

\(\displaystyle \frac{\partial F}{\partial x}=y^{-4}\)

\(\displaystyle \frac{\partial F}{\partial y}=-4\left(xy^{-5}+y \right)\)

And so we take:

\(\displaystyle \frac{\partial F}{\partial x}=y^{-4}\)

Now, what you need to do to get the solution is:

(a) Integrate this with respect to $x$, where the constant of integration is a function of $y$:

\(\displaystyle F(x,y)=\int y^{-4}\,dx+g(y)\)

(b) Then to determine $g(y)$, take the partials of both sides and then solve for $g'(y)$.

(c) Integrate $g'(y)$, omitting the constant of integration (as the implicit solution already contains an arbitrary constant).

(d) Replace $g(y)$ in the implicit solution you found in part(a).

Solving this for $x$ will give you an explicit solution (the one given by your textbook).
 

find_the_fun

Active member
Feb 1, 2012
166
You cannot treat this equation as linear, since we cannot write in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

...
Sorry I don't follow. I rewrote the equation \(\displaystyle \frac{dy}{dx} - \frac{y}{4(x+y^6)} = 0\) which is in the form and \(\displaystyle P(x)=\frac{-1}{4(x+y^6)}\) and \(\displaystyle Q(x)=0\) so isn't this equation in linear form?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Your function $P$ is a function of both $x$ and $y$. So the ODE is non-linear. :D
 

chisigma

Well-known member
Feb 13, 2012
1,704
You cannot treat this equation as linear, since we cannot write in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

We can however, use an integrating factor to get an equivalent exact equation.

We are given:

\(\displaystyle y\,dx-4\left(x+y^6 \right)\,dy=0\)

We see that:

\(\displaystyle \frac{\partial}{\partial y}(y)=1\)

\(\displaystyle \frac{\partial}{\partial x}\left(-4\left(x+y^6 \right) \right)=-4\)

Thus, the equation is inexact. So we consider:

\(\displaystyle \frac{\frac{\partial}{\partial x}\left(-4\left(x+y^6 \right) \right)-\frac{\partial}{\partial y}(y)}{y}=-\frac{5}{y}\)

So our integrating factor is:

\(\displaystyle \mu(y)=e^{\int -\frac{5}{y}\,dy}=y^{-5}\)

And thus our ODE becomes (observing that we are eliminating the trivial solution $y\equiv0$):

\(\displaystyle y^{-4}\,dx-4\left(xy^{-5}+y \right)\,dy=0\)

And we can easily see that this equation is exact. We know that the solution must therefore be given implicitly by:

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

And because the equation is exact, we know:

\(\displaystyle \frac{\partial F}{\partial x}=y^{-4}\)

\(\displaystyle \frac{\partial F}{\partial y}=-4\left(xy^{-5}+y \right)\)

And so we take:

\(\displaystyle \frac{\partial F}{\partial x}=y^{-4}\)

Now, what you need to do to get the solution is:

(a) Integrate this with respect to $x$, where the constant of integration is a function of $y$:

\(\displaystyle F(x,y)=\int y^{-4}\,dx+g(y)\)

(b) Then to determine $g(y)$, take the partials of both sides and then solve for $g'(y)$.

(c) Integrate $g'(y)$, omitting the constant of integration (as the implicit solution already contains an arbitrary constant).

(d) Replace $g(y)$ in the implicit solution you found in part(a).

Solving this for $x$ will give you an explicit solution (the one given by your textbook).
The procedure You used to find an integrating factor is very interesting and I want to realize if I have understood correctly. A differential expression like $\displaystyle A(x,y)\ dx + B(x,y)\ dy$ is 'exact' if $\displaystyle A_{y} (x,y) = B_{x} (x,y)$. In the OP is $\displaystyle A(x,y)= y$ and $B(x,y)= - 4\ (x + y^{6})$ so that the differential expression isn't exact. The idea is to make it exact by multiplying A and B by an unknown function $\displaystyle \mu(x,y)$ so that is exact the expression...

$\displaystyle A(x,y)\ \mu(x,y)\ dx + B (x,y)\ \mu (x,y)\ dy\ (1)$

... and that conducts to the PDE...

$\displaystyle A_{y} \ \mu + A\ \mu_{y} = B_{x}\ \mu + B\ \mu_{x} \implies \mu + y\ \mu_{y} = - 4\ \mu - 4\ (x+y^{6})\ \mu_{x}\ (2)$

The (2) seems not very comfortable, but if we suppose that $\mu$ is function of the y alone [i.e. $\displaystyle \mu_{x}=0$...] the (2) becomes the ODE...

$\displaystyle y\ \mu^{\ '} = -5\ \mu\ (3)$

... one solution of which is $\displaystyle \mu(y) = \frac{1}{y^{5}}$ that the integrating factor we searched...

It that 'all right'?...

Kind regards

$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The procedure You used to find an integrating factor is very interesting and I want to realize if I have understood correctly. A differential expression like $\displaystyle A(x,y)\ dx + B(x,y)\ dy$ is 'exact' if $\displaystyle A_{y} (x,y) = B_{x} (x,y)$. In the OP is $\displaystyle A(x,y)= y$ and $B(x,y)= - 4\ (x + y^{6})$ so that the differential expression isn't exact. The idea is to make it exact by multiplying A and B by an unknown function $\displaystyle \mu(x,y)$ so that is exact the expression...

$\displaystyle A(x,y)\ \mu(x,y)\ dx + B (x,y)\ \mu (x,y)\ dy\ (1)$

... and that conducts to the PDE...

$\displaystyle A_{y} \ \mu + A\ \mu_{y} = B_{x}\ \mu + B\ \mu_{x} \implies \mu + y\ \mu_{y} = - 4\ \mu - 4\ (x+y^{6})\ \mu_{x}\ (2)$

The (2) seems not very comfortable, but if we suppose that $\mu$ is function of the y alone [i.e. $\displaystyle \mu_{x}=0$...] the (2) becomes the ODE...

$\displaystyle y\ \mu^{\ '} = -5\ \mu\ (3)$

... one solution of which is $\displaystyle \mu(y) = \frac{1}{y^{5}}$ that the integrating factor we searched...

It that 'all right'?...

Kind regards

$\chi$ $\sigma$
Hello chisigma,

It looks like you have the gist of it quite well. :D

I will paraphrase from my old ODE textbook (Fundamentals of Differential Equations, 3rd Edition, by Nagle/Saff), the following information regarding integrating factors of inexact equations...

If the equation:

(1) \(\displaystyle M(x,y)dx+N(x,y)dy=0\)

is not exact, but the equation:

(2) \(\displaystyle \mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)dy=0\)

which results from multiplying equation (1) by the function $\mu(x,y)$, is exact, then $\mu(x,y)$ is called an integrating factor of the equation (1).

How do we find an integrating factor? If $\mu(x,y)$ is an integrating factor of (1) with continuous first partial derivatives, then testing (2) for exactness, we must have:

\(\displaystyle \frac{\partial}{\partial y}\left(\mu(x,y)M(x,y) \right)=\frac{\partial}{\partial x}\left(\mu(x,y)N(x,y) \right)\)

By use of the product rule, this reduces to the equation:

(3) \(\displaystyle M\frac{\partial \mu}{\partial y}-N\frac{\partial \mu}{\partial x}=\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)\mu\)

But solving the partial differential equation (3) for $\mu$ is usually more difficult than solving the original equation (1). There are, however, two important exceptions.

Let's assume equation (1) has an integrating factor that depends only on $x$, that is, $\mu=\mu(x)$. In this case, equation (3) reduces to the separable equation:

(4) \(\displaystyle \frac{d\mu}{dx}=\left(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N} \right)\mu\)

where \(\displaystyle \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is just a function of $x$.

In a similar fashion, if equation (1) has an integrating factor that depends only on $y$, then equation (3) reduces to the separable equation:

(5) \(\displaystyle \frac{d\mu}{dx}=\left(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} \right)\mu\)

where \(\displaystyle \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is just a function of $y$.

We can solve these separable equations (4) and (5) to obtain for (1) the integrating factors:

If \(\displaystyle \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is continuous and depends only on $x$, then:

(6) \(\displaystyle \mu(x)=\exp\left(\int\left(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N} \right)\,dx \right)\)

is an integrating factor for (1).

If \(\displaystyle \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is continuous and depends only on $y$, then:

(7) \(\displaystyle \mu(y)=\exp\left(\int\left(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} \right)\,dy \right)\)

is an integrating factor for (1).

There are many differential equations that are not covered by this method, but for which an integrating factor nevertheless exists. The major difficulty, however, is in finding an explicit formula for these integrating factors, which in general will depend on both $x$ and $y$.