Internal
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[7697] Book
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miscellaneous
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section
1.0 Problem
number
:
5 Date
solved
:
Monday, October 21, 2024 at 03:57:23 PM CAS
classification
:
[_quadrature]
Solve
\begin{align*} y^{\prime }&=y+1 \end{align*}
1.5.1 Solved as first order quadrature ode
Time used: 0.070 (sec)
Integrating gives
\begin{align*} \int \frac {1}{y +1}d y &= dx\\ \ln \left (y +1\right )&= x +c_1 \end{align*}
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.
\begin{align*} y = -1 \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
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 \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), 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 \(N\).
Where \(f(x)\) 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 \(x\) gives
Where \(c_1\) is constant of integration. Substituting result found above for \(f(x)\) into equation
(3) gives \(\phi \)
\[
\phi = {\mathrm e}^{-x} y +{\mathrm e}^{-x}+ c_1
\]
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
\[
c_1 = {\mathrm e}^{-x} y +{\mathrm e}^{-x}
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)