Internal problem ID [6211]
Internal file name [OUTPUT/5459_Sunday_June_05_2022_03_39_17_PM_43009656/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: 1(i).
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`], _rational, _Bernoulli]
\[ \boxed {x^{2} y^{\prime }-2 y x -y^{2}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \left (y +2 x \right )}{x^{2}}\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 \(M=y \left (y +2 x \right )\) and \(N=x^{2}\) 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 &= u^{2}+2 u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )^{2}+u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {u \left (x \right )^{2}+u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -u \left (x \right )^{2}-u \left (x \right ) = 0 \] 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 (u +1\right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u \left (u +1\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (u +1\right )} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u \left (u +1\right )} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (u \right )-\ln \left (u +1\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u \right )-\ln \left (u +1\right )} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {u}{u +1} &= 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 \[ y = -\frac {x^{2} c_{3}}{c_{3} x -1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{2} c_{3}}{c_{3} x -1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{2} c_{3}}{c_{3} x -1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-2 y x -y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y x +y^{2}}{x^{2}} \end {array} \]
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: 0.0 (sec). Leaf size: 15
dsolve(x^2*diff(y(x),x)=y(x)^2+2*x*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2}}{-x +c_{1}} \]
✓ Solution by Mathematica
Time used: 0.141 (sec). Leaf size: 23
DSolve[x^2*y'[x]==y[x]^2+2*x*y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x^2}{x-c_1} \\ y(x)\to 0 \\ \end{align*}