1.3 problem 3

Internal problem ID [7047]
Internal file name [OUTPUT/6033_Sunday_June_05_2022_04_14_40_PM_97928902/index.tex]

Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 3.
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]

\[ \boxed {y^{\prime }-\frac {\sec \left (x \right ) \left (\sin \left (y\right )+y\right )}{x}=0} \]

1.3.1 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\sec \left (x \right ) \left (\sin \left (y \right )+y \right )}{x} \end {align*}

Where \(f(x)=\frac {\sec \left (x \right )}{x}\) and \(g(y)=\sin \left (y \right )+y\). Integrating both sides gives \begin{align*} \frac {1}{\sin \left (y \right )+y} \,dy &= \frac {\sec \left (x \right )}{x} \,d x \\ \int { \frac {1}{\sin \left (y \right )+y} \,dy} &= \int {\frac {\sec \left (x \right )}{x} \,d x} \\ \int _{}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a}&=\int \frac {\sec \left (x \right )}{x}d x +c_{1} \\ \end{align*} Which results in \[ \int _{}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a}=\int \frac {\sec \left (x \right )}{x}d x +c_{1} \] The solution is \[ \int _{}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} -\left (\int \frac {\sec \left (x \right )}{x}d x \right )-c_{1} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} -\left (\int \frac {\sec \left (x \right )}{x}d x \right )-c_{1} &= 0 \\ \end{align*}

Figure 6: Slope field plot

Verification of solutions

\[ \int _{}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} -\left (\int \frac {\sec \left (x \right )}{x}d x \right )-c_{1} = 0 \] Verified OK.

1.3.2 Solving as first order ode lie symmetry lookup ode

Writing the ode as \begin {align*} y^{\prime }&=\frac {\sec \left (x \right ) \left (\sin \left (y \right )+y \right )}{x}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}

The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}

The type of this ode is known. It is of type separable. Therefore we do not need to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)

Table 5: Lie symmetry infinitesimal lookup table for known first order ODE’s

ODE class	Form	\(\xi \)	\(\eta \)
linear ode	\(y'=f(x) y(x) +g(x)\)	\(0\)	\(e^{\int fdx}\)
separable ode	\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \)	\(\frac {1}{f}\)	\(0\)
quadrature ode	\(y^{\prime }=f\left ( x\right ) \)	\(0\)	\(1\)
quadrature ode	\(y^{\prime }=g\left ( y\right ) \)	\(1\)	\(0\)
homogeneous ODEs of Class A	\(y^{\prime }=f\left ( \frac {y}{x}\right ) \)	\(x\)	\(y\)
homogeneous ODEs of Class C	\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\)	\(1\)	\(-\frac {b}{c}\)
homogeneous class D	\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \)	\(x^{2}\)	\(xy\)
First order special form ID 1	\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \)	\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\)	\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\)
polynomial type ode	\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\)	\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\)	\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\)
Bernoulli ode	\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\)	\(0\)	\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\)
Reduced Riccati	\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\)	\(0\)	\(e^{-\int f_{1}dx}\)

The above table shows that \begin {align*} \xi \left (x,y\right ) &=\frac {x}{\sec \left (x \right )}\\ \tag {A1} \eta \left (x,y\right ) &=0 \end {align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\eta =0\) then in this special case \begin {align*} R = y \end {align*}

\(S\) is found from \begin {align*} S &= \int { \frac {1}{\xi }} dx\\ &= \int { \frac {1}{\frac {x}{\sec \left (x \right )}}} dx \end {align*}

Which results in \begin {align*} S&= \int \frac {\sec \left (x \right )}{x}d x \end {align*}

1.3.3 Solving as exact ode

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 {1}{\sin \left (y \right )+y}\right )\mathop {\mathrm {d}y} &= \left (\frac {\sec \left (x \right )}{x}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {\sec \left (x \right )}{x}\right )\mathop {\mathrm {d}x} + \left (\frac {1}{\sin \left (y \right )+y}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}

Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -\frac {\sec \left (x \right )}{x}\\ N(x,y) &= \frac {1}{\sin \left (y \right )+y} \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 {\sec \left (x \right )}{x}\right )\\ &= 0 \end {align*}

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (\frac {1}{\sin \left (y \right )+y}\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 {\sec \left (x \right )}{x}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \int _{}^{x}-\frac {\sec \left (\textit {\_a} \right )}{\textit {\_a}}d \textit {\_a}+ 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 \(\frac {\partial \phi }{\partial y} = \frac {1}{\sin \left (y \right )+y}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {1}{\sin \left (y \right )+y} = 0+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = \frac {1}{\sin \left (y \right )+y} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{\sin \left (y \right )+y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= \int _{0}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a}+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \int _{}^{x}-\frac {\sec \left (\textit {\_a} \right )}{\textit {\_a}}d \textit {\_a} +\int _{0}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a}+ c_{1} \] 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 \[ c_{1} = \int _{}^{x}-\frac {\sec \left (\textit {\_a} \right )}{\textit {\_a}}d \textit {\_a} +\int _{0}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}-\frac {\sec \left (\textit {\_a} \right )}{\textit {\_a}}d \textit {\_a} +\int _{0}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} &= c_{1} \\ \end{align*}

Figure 7: Slope field plot

Verification of solutions

\[ \int _{}^{x}-\frac {\sec \left (\textit {\_a} \right )}{\textit {\_a}}d \textit {\_a} +\int _{0}^{y}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} = c_{1} \] Verified OK.

1.3.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\sec \left (x \right ) \left (\sin \left (y\right )+y\right )}{x}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sec \left (x \right ) \left (\sin \left (y\right )+y\right )}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sin \left (y\right )+y}=\frac {\sec \left (x \right )}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sin \left (y\right )+y}d x =\int \frac {\sec \left (x \right )}{x}d x +c_{1} \\ \bullet & {} & \textrm {Cannot compute integral}\hspace {3pt} \\ {} & {} & \int \frac {y^{\prime }}{\sin \left (y\right )+y}d x =\int \frac {\sec \left (x \right )}{x}d x +c_{1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 26

dsolve(diff(y(x),x) = sec(x)*(sin(y(x))+y(x))/x,y(x), singsol=all)
 

\[ \int \frac {\sec \left (x \right )}{x}d x -\left (\int _{}^{y \left (x \right )}\frac {1}{\sin \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} \right )+c_{1} = 0 \]

✓ Solution by Mathematica

Time used: 1.312 (sec). Leaf size: 41

DSolve[y'[x]== Sec[x]*(Sin[y[x]]+y[x])/x,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1]+\sin (K[1])}dK[1]\&\right ]\left [\int _1^x\frac {\sec (K[2])}{K[2]}dK[2]+c_1\right ] \]