Internal
problem
ID
[14790] Book
:
INTRODUCTORY
DIFFERENTIAL
EQUATIONS.
Martha
L.
Abell,
James
P.
Braselton.
Fourth
edition
2014.
ElScAe.
2014 Section
:
Chapter
2.
First
Order
Equations.
Exercises
2.1,
page
32 Problem
number
:
14
(b) Date
solved
:
Friday, October 11, 2024 at 05:12:32 PM CAS
classification
:
[_quadrature]
The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is
\[
\{-5<y <5\}
\]
But the point \(y_0 = 5\) is not inside this domain. Hence existence and
uniqueness theorem does not apply. Solution exists but no guarantee that unique solution
exists.
3.17.2 Solved as first order quadrature ode
Time used: 0.069 (sec)
Since the ode has the form \(y^{\prime }=f(y)\) and initial conditions \(\left (t_0,y_0\right ) \) are given such that they satisfy the ode
itself, then we can write
for \(y\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
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
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial t}\), 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 now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\)
and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(t\) 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
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparable<-separable successful`