Internal problem ID [2678]
Internal file name [OUTPUT/2170_Sunday_June_05_2022_02_51_36_AM_30877140/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 1, First-Order Differential Equations. Section 1.8, Change of Variables. page
79
Problem number: Problem 22.
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]
\[ \boxed {y^{\prime } x -x \tan \left (\frac {y}{x}\right )-y=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x \tan \left (\frac {y}{x}\right )+y}{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 \(M=x \tan \left (\frac {y}{x}\right )+y\) 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 &= \tan \left (u \right )+u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\tan \left (u \left (x \right )\right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\tan \left (u \left (x \right )\right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -\tan \left (u \left (x \right )\right ) = 0 \] 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 {\tan \left (u \right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\tan \left (u \right )\). Integrating both sides gives \begin{align*} \frac {1}{\tan \left (u \right )} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\tan \left (u \right )} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (\sin \left (u \right )\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sin \left (u \right ) &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sin \left (u \right ) &= c_{3} x \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = x \arcsin \left (c_{3} {\mathrm e}^{c_{2}} x \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= x \arcsin \left (c_{3} {\mathrm e}^{c_{2}} x \right ) \\ \end{align*}
Verification of solutions
\[ y = x \arcsin \left (c_{3} {\mathrm e}^{c_{2}} x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -x \tan \left (\frac {y}{x}\right )-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x \tan \left (\frac {y}{x}\right )+y}{x} \end {array} \]
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: 10
dsolve(x*diff(y(x),x)=x*tan(y(x)/x)+y(x),y(x), singsol=all)
\[ y \left (x \right ) = \arcsin \left (c_{1} x \right ) x \]
✓ Solution by Mathematica
Time used: 4.357 (sec). Leaf size: 19
DSolve[x*y'[x]==x*Tan[y[x]/x]+y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x \arcsin \left (e^{c_1} x\right ) \\ y(x)\to 0 \\ \end{align*}