4.18 problem Problem 27

Internal problem ID [2682]
Internal file name [OUTPUT/2174_Sunday_June_05_2022_02_51_53_AM_7818457/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 27.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"

Maple gives the following as the ode type

[[_homogeneous, `class A`], _dAlembert]

\[ \boxed {y^{\prime }-\frac {y-\sqrt {x^{2}+y^{2}}}{x}=0} \] With initial conditions \begin {align*} [y \left (3\right ) = 4] \end {align*}

4.18.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\frac {-y +\sqrt {x^{2}+y^{2}}}{x} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=4\) is \[ \{x <0\boldsymbol {\lor }0

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=4\) is \[ \{x <0\boldsymbol {\lor }0

4.18.2 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-y +\sqrt {x^{2}+y^{2}}}{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 \(M=y -\sqrt {x^{2}+y^{2}}\) 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 -\sqrt {u^{2}+1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= -\frac {\sqrt {u \left (x \right )^{2}+1}}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )+\frac {\sqrt {u \left (x \right )^{2}+1}}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x +\sqrt {u \left (x \right )^{2}+1} = 0 \] 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 {\sqrt {u^{2}+1}}{x} \end {align*}

Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\sqrt {u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\sqrt {u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\sqrt {u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \operatorname {arcsinh}\left (u \right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \operatorname {arcsinh}\left (u \left (x \right )\right )+\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \operatorname {arcsinh}\left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0 \] Substituting initial conditions and solving for \(c_{2}\) gives \(c_{2} = \operatorname {arcsinh}\left (\frac {4}{3}\right )+\ln \left (3\right )\). Hence the solution becomes

Summary

The solution(s) found are the following \begin{align*} \tag{1} \operatorname {arcsinh}\left (\frac {y}{x}\right )+\ln \left (x \right )-\operatorname {arcsinh}\left (\frac {4}{3}\right )-\ln \left (3\right ) &= 0 \\ \end{align*}

Verification of solutions

\[ \operatorname {arcsinh}\left (\frac {y}{x}\right )+\ln \left (x \right )-\operatorname {arcsinh}\left (\frac {4}{3}\right )-\ln \left (3\right ) = 0 \] Verified OK.

4.18.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {y-\sqrt {x^{2}+y^{2}}}{x}=0, y \left (3\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y-\sqrt {x^{2}+y^{2}}}{x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (3\right )=4 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous G 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful 
<- homogeneous successful`
 

✓ Solution by Maple

Time used: 0.375 (sec). Leaf size: 21

dsolve([diff(y(x),x)=(y(x)-sqrt(x^2+y(x)^2))/x,y(3) = 4],y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= \frac {x^{2}}{2}-\frac {1}{2} \\ y \left (x \right ) &= -\frac {x^{2}}{18}+\frac {9}{2} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.248 (sec). Leaf size: 29

DSolve[{y'[x]==(y[x]-Sqrt[x^2+y[x]^2])/x,{y[3]==4}},y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {9}{2}-\frac {x^2}{18} \\ y(x)\to \frac {1}{2} \left (x^2-1\right ) \\ \end{align*}