Internal problem ID [2671]
Internal file name [OUTPUT/2163_Sunday_June_05_2022_02_51_10_AM_21063386/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 15.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_separable]
\[ \boxed {y \left (x^{2}-y^{2}\right )-x \left (x^{2}-y^{2}\right ) y^{\prime }=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {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 \(M=y\) 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 &= u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= 0 \end {align*}
Or \[ u^{\prime }\left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). Integrating both sides gives \begin {align*} u \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{2} \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = c_{2} x \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x \\ \end{align*}
Verification of solutions
\[ y = c_{2} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x^{2}-y^{2}\right )-x \left (x^{2}-y^{2}\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} x \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 17
dsolve(y(x)*(x^2-y(x)^2)-x*(x^2-y(x)^2)*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x \\ y \left (x \right ) &= x \\ y \left (x \right ) &= c_{1} x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 33
DSolve[y[x]*(x^2-y[x]^2)-x*(x^2-y[x]^2)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x \\ y(x)\to x \\ y(x)\to c_1 x \\ y(x)\to -x \\ y(x)\to x \\ \end{align*}