Internal problem ID [5767]
Internal file name [OUTPUT/5015_Sunday_June_05_2022_03_17_25_PM_7693094/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.2 Homogeneous equations
problems. page 12
Problem number: 19.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "bernoulli", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _Bernoulli]
\[ \boxed {y^{\prime }-\frac {x}{y}-\frac {y}{x}=0} \] With initial conditions \begin {align*} [y \left (-1\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {x^{2}+y^{2}}{x y} \end {align*}
\(f(x,y)\) is not defined at \(y = 0\) therefore existence and uniqueness theorem do not apply.
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2}+y^{2}}{x y}\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 \(M=x^{2}+y^{2}\) and \(N=x y\) are both homogeneous and of the same order \(n=2\). 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 &= \frac {1}{u}+u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {1}{u \left (x \right ) x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {1}{u \left (x \right ) x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right ) x -1 = 0 \] 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 {1}{u x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {1}{u}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {u^{2}}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {u \left (x \right )^{2}}{2}-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} = 0 \] Substituting initial conditions and solving for \(c_{2}\) gives \(c_{2} = -i \pi \). Hence the solution becomes
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )+i \pi &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )+i \pi = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {x}{y}-\frac {y}{x}=0, y \left (-1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x}{y}+\frac {y}{x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 34
dsolve([diff(y(x),x)=x/y(x)+y(x)/x,y(-1) = 0],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {2 \ln \left (x \right )-2 i \pi }\, x \\ y \left (x \right ) &= -\sqrt {2 \ln \left (x \right )-2 i \pi }\, x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.19 (sec). Leaf size: 48
DSolve[{y'[x]==x/y[x]+y[x]/x,{y[-1]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {2} x \sqrt {\log (x)-i \pi } \\ y(x)\to \sqrt {2} x \sqrt {\log (x)-i \pi } \\ \end{align*}