1.2 problem 2
Internal
problem
ID
[7694]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
2
Date
solved
:
Monday, October 21, 2024 at 03:56:59 PM
CAS
classification
:
[_separable]
Solve
\begin{align*} y^{\prime }&=x \left (\cos \left (y\right )+y\right ) \end{align*}
1.2.1 Solved as first order separable ode
Time used: 0.233 (sec)
The ode \(y^{\prime } = x \left (\cos \left (y\right )+y\right )\) is separable as it can be written as
\begin{align*} y^{\prime }&= x \left (\cos \left (y\right )+y\right )\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= x\\ g(y) &= \cos \left (y \right )+y \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{\cos \left (y \right )+y}\,dy} &= \int { x \,dx}\\ \int _{}^{y}\frac {1}{\cos \left (\tau \right )+\tau }d \tau = \frac {x^{2}}{2}+c_1 \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is
zero, since we had to divide by this above. Solving \(g(y)=0\) or \(\cos \left (y \right )+y=0\) for \(y\) gives
\begin{align*} y&=\operatorname {RootOf}\left (\cos \left (\textit {\_Z} \right )+\textit {\_Z} \right ) \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(\operatorname {RootOf}\left (\cos \left (\textit {\_Z} \right )+\textit {\_Z} \right )\) will not be used
Figure 3: Slope field plot
\(y^{\prime } = x \left (\cos \left (y\right )+y\right )\)
1.2.2 Solved as first order Exact ode
Time used: 0.214 (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*} \mathop {\mathrm {d}y} &= \left (x \left (\cos \left (y \right )+y \right )\right )\mathop {\mathrm {d}x}\\ \left (-x \left (\cos \left (y \right )+y \right )\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -x \left (\cos \left (y \right )+y \right )\\ N(x,y) &= 1 \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 (-x \left (\cos \left (y \right )+y \right )\right )\\ &= x \left (\sin \left (y \right )-1\right ) \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}
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
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( -x \left (-\sin \left (y \right )+1\right )\right ) - \left (0 \right ) \right ) \\ &=x \left (\sin \left (y \right )-1\right ) \end{align*}
Since \(A\) depends on \(y\) , it can not be used to obtain an integrating factor. We will now try a
second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {1}{x \left (\cos \left (y \right )+y \right )}\left ( \left ( 0\right ) - \left (-x \left (-\sin \left (y \right )+1\right ) \right ) \right ) \\ &=\frac {\sin \left (y \right )-1}{\cos \left (y \right )+y} \end{align*}
Since \(B\) does not depend on \(x\) , it can be used to obtain an integrating factor. Let the
integrating factor be \(\mu \) . Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}y}} \\ &= e^{\int \frac {\sin \left (y \right )-1}{\cos \left (y \right )+y}\mathop {\mathrm {d}y} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-\ln \left (\cos \left (y \right )+y \right ) } \\ &= \frac {1}{\cos \left (y \right )+y} \end{align*}
\(M\) 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 \(N\) .
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{\cos \left (y \right )+y}\left (-x \left (\cos \left (y \right )+y \right )\right ) \\ &= -x \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{\cos \left (y \right )+y}\left (1\right ) \\ &= \frac {1}{\cos \left (y \right )+y} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be
solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-x\right ) + \left (\frac {1}{\cos \left (y \right )+y}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {M}\mathop {\mathrm {d}x} \\
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -x\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -\frac {x^{2}}{2}+ 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 \(\frac {\partial \phi }{\partial y} = \frac {1}{\cos \left (y \right )+y}\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {1}{\cos \left (y \right )+y} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = \frac {1}{\cos \left (y \right )+y}
\]
Integrating the above w.r.t \(y\)
gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{\cos \left (y \right )+y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \int _{}^{y}\frac {1}{\cos \left (\tau \right )+\tau }d \tau + c_1 \\
\end{align*}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into
equation (3) gives \(\phi \)
\[
\phi = -\frac {x^{2}}{2}+\int _{}^{y}\frac {1}{\cos \left (\tau \right )+\tau }d \tau + c_1
\]
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
\[
c_1 = -\frac {x^{2}}{2}+\int _{}^{y}\frac {1}{\cos \left (\tau \right )+\tau }d \tau
\]
Figure 4: Slope field plot
\(y^{\prime } = x \left (\cos \left (y\right )+y\right )\)
1.2.3 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=x \left (\cos \left (y\right )+y\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x \left (\cos \left (y\right )+y\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\cos \left (y\right )+y}=x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\cos \left (y\right )+y}d x =\int x d x +\mathit {C1} \\ \bullet & {} & \textrm {Cannot compute integral}\hspace {3pt} \\ {} & {} & \int \frac {y^{\prime }}{\cos \left (y\right )+y}d x =\frac {x^{2}}{2}+\mathit {C1} \end {array} \]
1.2.4 Maple trace
Methods for first order ODEs:
1.2.5 Maple dsolve solution
Solving time : 0.003
(sec)
Leaf size : 23
dsolve ( diff ( y ( x ), x ) = x*( cos ( y ( x ))+ y ( x )),
y(x),singsol=all)
\[
\frac {x^{2}}{2}-\left (\int _{}^{y}\frac {1}{\cos \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} \right )+c_1 = 0
\]
1.2.6 Mathematica DSolve solution
Solving time : 0.625
(sec)
Leaf size : 33
DSolve [{ D [ y [ x ], x ] == x*( Cos [y[x]]+y[x]),{}},
y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{\cos (K[1])+K[1]}dK[1]\&\right ]\left [\frac {x^2}{2}+c_1\right ]
\]