Internal
problem
ID
[3827] Book
:
Differential
equations
for
engineers
by
Wei-Chau
XIE,
Cambridge
Press
2010 Section
:
Chapter
2.
First-Order
and
Simple
Higher-Order
Differential
Equations.
Page
78 Problem
number
:
32 Date
solved
:
Thursday, October 17, 2024 at 04:50:20 AM CAS
classification
:
[_rational, [_1st_order, `_with_symmetry_[F(x),G(x)]`], [_Abel, `2nd type`, `class B`]]
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (3 x^{2} y -x\right )\\ &= 6 x y -1 \end{align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\)
for now so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{x^{2}}\left (x^{2}+2 x +y\right ) \\ &= \frac {x^{2}+2 x +y}{x^{2}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{x^{2}}\left (3 x^{2} y -x\right ) \\ &= \frac {3 x y -1}{x} \end{align*}
Now a modified ODE is ontained from the original ODE, which is exact and can be solved.
The modified ODE is
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives
But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and
combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2}+2 x +y \left (x \right )+\left (3 x^{2} y \left (x \right )-x \right ) \left (\frac {d}{d x}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-x^{2}-2 x -y \left (x \right )}{3 x^{2} y \left (x \right )-x} \end {array} \]
1.32.3 Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryingChinidifferentialorder: 1; looking for linear symmetriestryingexact<-exact successful`