Internal problem ID [3155]
Internal file name [OUTPUT/2647_Sunday_June_05_2022_08_38_04_AM_4756193/index.tex
]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 10.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeD", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _dAlembert]
\[ \boxed {x \cos \left (\frac {y}{x}\right )^{2}-y+y^{\prime } x=0} \]
Writing the ode as \begin {align*} y^{\prime }&=-\cos \left (\frac {y}{x}\right )^{2}+\frac {y}{x}\tag {A} \end {align*}
The given ode has the form\begin {equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}}\tag {1} \end {equation} Where \(b\) is scalar and \(g\left ( x\right ) \) is function of \(x\) and \(n,m\) are integers. The solution is given in Kamke page 20. Using the substitution \(y\left ( x\right ) =u\left ( x\right ) x\) then\[ \frac {dy}{dx}=\frac {du}{dx}x+u \] Hence the given ode becomes\begin {align} \frac {du}{dx}x+u & =u+g\left ( x\right ) f\left ( bu\right ) ^{\frac {n}{m}}\nonumber \\ u^{\prime } & =\frac {1}{x}g\left ( x\right ) f\left ( bu\right ) ^{\frac {n}{m}}\tag {2} \end {align}
The above ode is always separable. This is easily solved for \(u\) assuming the integration can be resolved, and then the solution to the original ode becomes \(y=ux\). Comapring the given ode (A) with the form (1) shows that \begin {align*} g \left (x \right )&=-1\\ b&=1\\ f \left (\frac {b x}{y}\right )&=\cos \left (\frac {y}{x}\right ) \end {align*}
Substituting the above in (2) results in the \(u(x)\) ode as \begin {align*} u^{\prime }\left (x \right ) = -\frac {\cos \left (u \left (x \right )\right )^{2}}{x} \end {align*}
Which is now solved as separable In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {\cos \left (u \right )^{2}}{x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\cos \left (u \right )^{2}\). Integrating both sides gives \begin{align*} \frac {1}{\cos \left (u \right )^{2}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\cos \left (u \right )^{2}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \tan \left (u \right )&=-\ln \left (x \right )+c_{1} \\ \end{align*} The solution is \[ \tan \left (u \left (x \right )\right )+\ln \left (x \right )-c_{1} = 0 \] Therefore the solution is found using \(y = u x\). Hence \begin {align*} \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{1} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{1} = 0 \] Verified OK.
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} x \cos \left (u \left (x \right )\right )^{2}-u \left (x \right ) x +\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {\cos \left (u \right )^{2}}{x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\cos \left (u \right )^{2}\). Integrating both sides gives \begin{align*} \frac {1}{\cos \left (u \right )^{2}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\cos \left (u \right )^{2}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \tan \left (u \right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \tan \left (u \left (x \right )\right )+\ln \left (x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0\\ \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \tan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0 \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=-\frac {x \cos \left (\frac {y}{x}\right )^{2}-y}{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 homogeneous Type D
. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
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 ) &=x^{2}\\ \tag {A1} \eta \left (x,y\right ) &=x y \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 \(S(R)\). Unable to determine \(R\). Terminating
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \cos \left (\frac {y}{x}\right )^{2}-y+y^{\prime } x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-x \cos \left (\frac {y}{x}\right )^{2}+y}{x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
dsolve((x*cos(y(x)/x)^2-y(x))+x*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = -\arctan \left (\ln \left (x \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.5 (sec). Leaf size: 37
DSolve[(x*Cos[y[x]/x]^2-y[x])+x*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x \arctan (-\log (x)+2 c_1) \\ y(x)\to -\frac {\pi x}{2} \\ y(x)\to \frac {\pi x}{2} \\ \end{align*}