2.2.5 Internal
problem
ID
[20971]Book
:
Ordinary
Differential
Equations.
By
Wolfgang
Walter.
Graduate
texts
in
Mathematics.
Springer.
NY.
QA372.W224
1998 Section
:
Chapter
1.
First
order
equations:
Some
integrable
cases.
Excercises
XIII
at
page
24 Problem
number
:
(f)Date
solved
:
Saturday, November 29, 2025 at 01:16:07 AMCAS
classification
:
[_separable]
\begin{align*}
y^{\prime }&=\frac {\cos \left (x \right )}{\cos \left (y\right )^{2}} \\
y \left (\pi \right ) &= \frac {\pi }{4} \\
\end{align*}
This is non linear first order ODE. In canonical form it is written as
\begin{align*} y^{\prime } &= f(x,y)\\ &= \frac {\cos \left (x \right )}{\cos \left (y \right )^{2}} \end{align*}
The \(x\) domain of \(f(x,y)\) when \(y=\frac {\pi }{4}\) is
\[
\{-\infty <x <\infty \}
\]
And the point
\(x_0 = \pi \) is inside this domain. The
\(y\) domain of
\(f(x,y)\) when
\(x=\pi \) is
\[
\left \{y <\frac {1}{2} \pi +\pi \_Z357 \boldsymbol {\lor }\frac {1}{2} \pi +\pi \_Z357 <y\right \}
\]
And the
point
\(y_0 = \frac {\pi }{4}\) is inside this domain. Now we will look at the continuity of
\begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (\frac {\cos \left (x \right )}{\cos \left (y \right )^{2}}\right ) \\ &= \frac {2 \cos \left (x \right ) \sin \left (y \right )}{\cos \left (y \right )^{3}} \end{align*}
The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=\frac {\pi }{4}\) is
\[
\{-\infty <x <\infty \}
\]
And the point
\(x_0 = \pi \) is inside this domain. The
\(y\) domain of
\(\frac {\partial f}{\partial y}\) when
\(x=\pi \) is
\[
\left \{y <\frac {1}{2} \pi +\pi \_Z357 \boldsymbol {\lor }\frac {1}{2} \pi +\pi \_Z357 <y\right \}
\]
And the
point
\(y_0 = \frac {\pi }{4}\) is inside this domain. Therefore solution exists and is unique.
Time used: 0.171 (sec)
Solve
\begin{align*}
y^{\prime }&=\frac {\cos \left (x \right )}{\cos \left (y\right )^{2}} \\
y \left (\pi \right ) &= \frac {\pi }{4} \\
\end{align*}
The ode
\begin{equation}
y^{\prime } = \frac {\cos \left (x \right )}{\cos \left (y\right )^{2}}
\end{equation}
is separable as it can be written as
\begin{align*} y^{\prime }&= \frac {\cos \left (x \right )}{\cos \left (y\right )^{2}}\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= \cos \left (x \right )\\ g(y) &= \frac {1}{\cos \left (y \right )^{2}} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx} \\
\int { \cos \left (y \right )^{2}\,dy} &= \int { \cos \left (x \right ) \,dx} \\
\end{align*}
\[
\frac {\cos \left (y\right ) \sin \left (y\right )}{2}+\frac {y}{2}=\sin \left (x \right )+c_1
\]
Simplifying the above gives
\begin{align*}
\frac {\sin \left (2 y\right )}{4}+\frac {y}{2} &= \sin \left (x \right )+c_1 \\
\end{align*}
Solving for initial conditions the solution is
\begin{align*}
\frac {\sin \left (2 y\right )}{4}+\frac {y}{2} &= \sin \left (x \right )+\frac {1}{4}+\frac {\pi }{8} \\
\end{align*}
Solving
for
\(y\) gives
\begin{align*}
y &= \frac {\operatorname {RootOf}\left (2 \textit {\_Z} +2 \sin \left (\textit {\_Z} \right )-2-8 \sin \left (x \right )-\pi \right )}{2} \\
\end{align*}
Figure 2.27: Slope field \(y^{\prime } = \frac {\cos \left (x \right )}{\cos \left (y\right )^{2}}\) Summary of solutions found
\begin{align*}
y &= \frac {\operatorname {RootOf}\left (2 \textit {\_Z} +2 \sin \left (\textit {\_Z} \right )-2-8 \sin \left (x \right )-\pi \right )}{2} \\
\end{align*}
Time used: 0.088 (sec)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
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
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows
that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
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
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
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
\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]
Therefore
\begin{align*} \left (\cos \left (y \right )^{2}\right )\mathop {\mathrm {d}y} &= \left (\cos \left (x \right )\right )\mathop {\mathrm {d}x}\\ \left (-\cos \left (x \right )\right )\mathop {\mathrm {d}x} + \left (\cos \left (y \right )^{2}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -\cos \left (x \right )\\ N(x,y) &= \cos \left (y \right )^{2} \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following
condition is satisfied
\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\cos \left (x \right )\right )\\ &= 0 \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (\cos \left (y \right )^{2}\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\) , then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
\begin{align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end{align*}
Integrating (1) w.r.t. \(x\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\cos \left (x \right )\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -\sin \left (x \right )+ f(y) \\
\end{align*}
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
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y)
\end{equation}
But equation (2) says that
\(\frac {\partial \phi }{\partial y} = \cos \left (y \right )^{2}\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} \cos \left (y \right )^{2} = 0+f'(y)
\end{equation}
Solving equation (5) for
\( f'(y)\) gives
\[
f'(y) = \cos \left (y \right )^{2}
\]
Integrating the above
w.r.t
\(y\) gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \cos \left (y \right )^{2}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \frac {\cos \left (y \right ) \sin \left (y \right )}{2}+\frac {y}{2}+ c_1 \\
\end{align*}
Where
\(c_1\) is constant of integration. Substituting result found above for
\(f(y)\)
into equation (3) gives
\(\phi \) \[
\phi = -\sin \left (x \right )+\frac {\cos \left (y \right ) \sin \left (y \right )}{2}+\frac {y}{2}+ c_1
\]
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
\[
c_1 = -\sin \left (x \right )+\frac {\cos \left (y \right ) \sin \left (y \right )}{2}+\frac {y}{2}
\]
Simplifying the above gives
\begin{align*}
-\sin \left (x \right )+\frac {\sin \left (2 y\right )}{4}+\frac {y}{2} &= c_1 \\
\end{align*}
Solving for initial conditions the solution is
\begin{align*}
-\sin \left (x \right )+\frac {\sin \left (2 y\right )}{4}+\frac {y}{2} &= \frac {1}{4}+\frac {\pi }{8} \\
\end{align*}
Solving for
\(y\) gives
\begin{align*}
y &= \frac {\operatorname {RootOf}\left (2 \textit {\_Z} +2 \sin \left (\textit {\_Z} \right )-2-8 \sin \left (x \right )-\pi \right )}{2} \\
\end{align*}
Figure 2.28: Slope field \(y^{\prime } = \frac {\cos \left (x \right )}{\cos \left (y\right )^{2}}\) Summary of solutions found
\begin{align*}
y &= \frac {\operatorname {RootOf}\left (2 \textit {\_Z} +2 \sin \left (\textit {\_Z} \right )-2-8 \sin \left (x \right )-\pi \right )}{2} \\
\end{align*}
✓ Time used: 0.122 (sec). Leaf size: 23 ode := diff ( y ( x ), x ) = cos(x)/cos(y(x))^2;
ic :=[ y ( Pi ) = 1/4*Pi];
dsolve ([ ode , op ( ic )], y ( x ), singsol=all);
\[
y = \frac {\operatorname {RootOf}\left (2 \textit {\_Z} -\pi -2-8 \sin \left (x \right )+2 \sin \left (\textit {\_Z} \right )\right )}{2}
\]
Maple trace
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\cos \left (x \right )}{\cos \left (y \left (x \right )\right )^{2}}, y \left (\pi \right )=\frac {\pi }{4}\right ] \\ \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 {\cos \left (x \right )}{\cos \left (y \left (x \right )\right )^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \cos \left (y \left (x \right )\right )^{2}=\cos \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right ) \cos \left (y \left (x \right )\right )^{2}d x =\int \cos \left (x \right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\cos \left (y \left (x \right )\right ) \sin \left (y \left (x \right )\right )}{2}+\frac {y \left (x \right )}{2}=\sin \left (x \right )+\mathit {C1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\pi \right )=\frac {\pi }{4} \\ {} & {} & \frac {1}{4}+\frac {\pi }{8}=\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\frac {1}{4}+\frac {\pi }{8} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\frac {1}{4}+\frac {\pi }{8}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & \frac {\sin \left (2 y \left (x \right )\right )}{4}+\frac {y \left (x \right )}{2}=\sin \left (x \right )+\frac {1}{4}+\frac {\pi }{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & \frac {\sin \left (2 y \left (x \right )\right )}{4}+\frac {y \left (x \right )}{2}=\sin \left (x \right )+\frac {1}{4}+\frac {\pi }{8} \end {array} \]
✓ Time used: 0.227 (sec). Leaf size: 36 ode = D [ y [ x ], x ]== Cos[x]/Cos[y[x]]^2;
ic ={ y [ Pi ]== Pi /4}; DSolve [{ ode , ic }, y [ x ], x , IncludeSingularSolutions -> True ] \begin{align*} y(x)&\to \text {InverseFunction}\left [2 \left (\frac {\text {$\#$1}}{2}+\frac {1}{4} \sin (2 \text {$\#$1})\right )\&\right ]\left [\frac {1}{4} (8 \sin (x)+\pi +2)\right ] \end{align*}
✓ Time used: 3.051 (sec). Leaf size: 26 from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-cos(x)/cos(y(x))**2 + Derivative(y(x), x),0)
ics = {y(pi): pi/4}
dsolve ( ode , func = y ( x ), ics = ics ) \[
\frac {y{\left (x \right )}}{2} - \sin {\left (x \right )} + \frac {\sin {\left (y{\left (x \right )} \right )} \cos {\left (y{\left (x \right )} \right )}}{2} = \frac {1}{4} + \frac {\pi }{8}
\]