Internal
problem
ID
[18491] Book
:
Elementary
Differential
Equations.
By
R.L.E.
Schwarzenberger.
Chapman
and
Hall.
London.
First
Edition
(1969) Section
:
Chapter
3.
Solutions
of
first-order
equations.
Exercises
at
page
47 Problem
number
:
3
(i) Date
solved
:
Saturday, February 22, 2025 at 09:17:40 PM CAS
classification
:
[_separable]
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 t} &= \frac {\partial }{\partial t} \left (t^{2} x \left (x^{2}+3\right )\right )\\ &= 2 t x \left (x^{2}+3\right ) \end{align*}
Since \(\frac {\partial M}{\partial x} \neq \frac {\partial N}{\partial t}\), 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}{t^{2}}\left (3 t -1\right ) \\ &= \frac {3 t -1}{t^{2}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{t^{2}}\left (t^{2} x \left (x^{2}+3\right )\right ) \\ &= x \left (x^{2}+3\right ) \end{align*}
Now a modified ODE is ontained from the original ODE, which is exact and can be solved. The modified ODE is
Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function of both \(t\) and \(x\). Taking derivative of equation (3) w.r.t \(x\) 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 t^{2} x-x t +\left (3 t^{3} x^{2}+t^{3} x^{4}\right ) x^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=\frac {-3 t^{2} x+x t}{3 t^{3} x^{2}+t^{3} x^{4}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & x^{\prime } x \left (x^{2}+3\right )=-\frac {3 t -1}{t^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int x^{\prime } x \left (x^{2}+3\right )d t =\int -\frac {3 t -1}{t^{2}}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\left (x^{2}+3\right )^{2}}{4}=-\frac {1}{t}-3 \ln \left (t \right )+\mathit {C1} \end {array} \]
Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparable<-separable successful`