Internal problem ID [4466]
Internal file name [OUTPUT/3959_Sunday_June_05_2022_11_55_48_AM_90518544/index.tex
]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 2. Special types of differential equations of the first kind. Lesson
9
Problem number: Exact Differential equations. Exercise 9.16, page 79.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "separable", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {\sin \left (x \right ) \cos \left (y\right )+\cos \left (x \right ) \sin \left (y\right ) y^{\prime }=0} \] With initial conditions \begin {align*} \left [y \left (\frac {\pi }{4}\right ) = \frac {\pi }{4}\right ] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\frac {\sin \left (x \right ) \cos \left (y \right )}{\cos \left (x \right ) \sin \left (y \right )} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=\frac {\pi }{4}\) is \[
\left \{x <\frac {1}{2} \pi +\pi \_Z77 \boldsymbol {\lor }\frac {1}{2} \pi +\pi \_Z77 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=\frac {\pi }{4}\) is \[
\left \{x <\frac {1}{2} \pi +\pi \_Z77 \boldsymbol {\lor }\frac {1}{2} \pi +\pi \_Z77
Entering Exact first order ODE solver. (Form one type)
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 (-\frac {\sin \left (y \right )}{\cos \left (y \right )}\right )\mathop {\mathrm {d}y} &= \left (\frac {\sin \left (x \right )}{\cos \left (x \right )}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {\sin \left (x \right )}{\cos \left (x \right )}\right )\mathop {\mathrm {d}x} + \left (-\frac {\sin \left (y \right )}{\cos \left (y \right )}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -\frac {\sin \left (x \right )}{\cos \left (x \right )}\\ N(x,y) &= -\frac {\sin \left (y \right )}{\cos \left (y \right )} \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 (-\frac {\sin \left (x \right )}{\cos \left (x \right )}\right )\\ &= 0 \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (-\frac {\sin \left (y \right )}{\cos \left (y \right )}\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 -\frac {\sin \left (x \right )}{\cos \left (x \right )}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= \ln \left (\cos \left (x \right )\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 \(\frac {\partial \phi }{\partial y} = -\frac {\sin \left (y \right )}{\cos \left (y \right )}\).
Therefore equation (4) becomes \begin{equation}
\tag{5} -\frac {\sin \left (y \right )}{\cos \left (y \right )} = 0+f'(y)
\end{equation} Solving equation (5) for \( f'(y)\) gives \begin{align*}
f'(y) &= -\frac {\sin \left (y \right )}{\cos \left (y \right )} \\
&= -\tan \left (y \right )\\
\end{align*} Integrating the
above w.r.t \(y\) results in \begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\tan \left (y \right )\right ) \mathop {\mathrm {d}y} \\
f(y) &= \ln \left (\cos \left (y \right )\right )+ c_{1} \\
\end{align*} Where \(c_{1}\) is constant of integration. Substituting result found
above for \(f(y)\) into equation (3) gives \(\phi \) \[
\phi = \ln \left (\cos \left (x \right )\right )+\ln \left (\cos \left (y \right )\right )+ c_{1}
\] 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 new constant \(c_{1}\)
gives the solution as \[
c_{1} = \ln \left (\cos \left (x \right )\right )+\ln \left (\cos \left (y \right )\right )
\] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=\frac {\pi }{4}\) and \(y=\frac {\pi }{4}\)
in the above solution gives an equation to solve for the constant of integration.
\begin {align*} -\ln \left (2\right ) = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -\ln \left (2\right ) \end {align*}
Trying the constant \begin {align*} c_{1} = -\ln \left (2\right ) \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \ln \left (\cos \left (x \right )\right )+\ln \left (\cos \left (y \right )\right ) = -\ln \left (2\right ) \end {align*}
The constant \(c_{1} = -\ln \left (2\right )\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \ln \left (\cos \left (x \right )\right )+\ln \left (\cos \left (y\right )\right ) &= -\ln \left (2\right ) \\
\end{align*} Verification of solutions
\[
\ln \left (\cos \left (x \right )\right )+\ln \left (\cos \left (y\right )\right ) = -\ln \left (2\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\sin \left (x \right ) \cos \left (y\right )+\cos \left (x \right ) \sin \left (y\right ) y^{\prime }=0, y \left (\frac {\pi }{4}\right )=\frac {\pi }{4}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } \sin \left (y\right )}{\cos \left (y\right )}=-\frac {\sin \left (x \right )}{\cos \left (x \right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } \sin \left (y\right )}{\cos \left (y\right )}d x =\int -\frac {\sin \left (x \right )}{\cos \left (x \right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (\cos \left (y\right )\right )=\ln \left (\cos \left (x \right )\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\arccos \left (\frac {1}{\cos \left (x \right ) {\mathrm e}^{c_{1}}}\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\frac {\pi }{4}\right )=\frac {\pi }{4} \\ {} & {} & \frac {\pi }{4}=\arccos \left (\frac {\sqrt {2}}{{\mathrm e}^{c_{1}}}\right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\arccos \left (\frac {\sec \left (x \right )}{2}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\arccos \left (\frac {\sec \left (x \right )}{2}\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.141 (sec). Leaf size: 9
\[
y \left (x \right ) = \frac {\pi }{2}-\arcsin \left (\frac {\sec \left (x \right )}{2}\right )
\]
✓ Solution by Mathematica
Time used: 6.111 (sec). Leaf size: 12
\[
y(x)\to \arccos \left (\frac {\sec (x)}{2}\right )
\]
3.12.2 Solving as exact ode
3.12.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([sin(x)*cos(y(x))+cos(x)*sin(y(x))*diff(y(x),x)=0,y(1/4*Pi) = 1/4*Pi],y(x), singsol=all)
DSolve[{Sin[x]*Cos[y[x]]+Cos[x]*Sin[y[x]]*y'[x]==0,y[Pi/4]==Pi/4},y[x],x,IncludeSingularSolutions -> True]