Internal problem ID [6224]
Internal file name [OUTPUT/5472_Sunday_June_05_2022_03_40_15_PM_31766225/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.7. Homogeneous Equations. Page
28
Problem number: 7(d).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class A`], _dAlembert]
\[ \boxed {y^{\prime }-\frac {y \tan \left (\frac {y}{x}\right )}{x}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \tan \left (\frac {y}{x}\right )}{x}\tag {1} \end {align*}
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if \[ f(t^n x, t^n y)= t^n f(x,y) \] In this case, it can be seen that both \(M=y \tan \left (\frac {y}{x}\right )\) and \(N=x\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence \[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \] Applying the transformation \(y=ux\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= u \tan \left (u \right )\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right ) \tan \left (u \left (x \right )\right )-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {u \left (x \right ) \tan \left (u \left (x \right )\right )-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -u \left (x \right ) \tan \left (u \left (x \right )\right )+u \left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (\tan \left (u \right )-1\right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u \left (\tan \left (u \right )-1\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (\tan \left (u \right )-1\right )} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u \left (\tan \left (u \right )-1\right )} \,du} &= \int {\frac {1}{x} \,d x} \\ \int _{}^{u}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a}&=\ln \left (x \right )+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a}=\ln \left (x \right )+c_{2} \] The solution is \[ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a} -\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\tan \left (\textit {\_a} \right )-1\right )}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y \tan \left (\frac {y}{x}\right )}{x}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \tan \left (\frac {y}{x}\right )}{x} \end {array} \]
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: 27
dsolve(diff(y(x),x)=y(x)/x*tan(y(x)/x),y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (\ln \left (x \right )+c_{1} -\left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a} \left (-1+\tan \left (\textit {\_a} \right )\right )}d \textit {\_a} \right )\right ) x \]
✓ Solution by Mathematica
Time used: 1.796 (sec). Leaf size: 33
DSolve[y'[x]==y[x]/x*Tan[y[x]/x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{\frac {y(x)}{x}}\frac {1}{K[1] (\tan (K[1])-1)}dK[1]=\log (x)+c_1,y(x)\right ] \]