Internal
problem
ID
[9021] Book
:
First
order
enumerated
odes Section
:
section
1 Problem
number
:
37 Date
solved
:
Wednesday, March 05, 2025 at 07:14:50 AM CAS
classification
:
[_quadrature]
Solve
\begin{align*} x y^{\prime }&=1 \end{align*}
Solved as first order quadrature ode
Time used: 0.023 (sec)
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
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 \(f(y)\) 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
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = -\ln \left (x \right )+y+ c_1
\]
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
\[
c_1 = -\ln \left (x \right )+y
\]
Solving for \(y\) gives
\begin{align*}
y &= \ln \left (x \right )+c_1 \\
\end{align*}
Figure 2.43: Slope field \(x y^{\prime } = 1\)
Summary of solutions found
\begin{align*}
y &= \ln \left (x \right )+c_1 \\
\end{align*}