2.1.30 Problem 31

Internal problem ID [8742]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 31
Date solved : Wednesday, March 05, 2025 at 06:43:32 AM
CAS classification : [[_homogeneous, `class A`], _rational, _Riccati]

Solve

\begin{align*} y^{\prime }&=\frac {5 x^{2}-x y+y^{2}}{x^{2}} \end{align*}

Solved as first order homogeneous class A ode

Time used: 0.349 (sec)

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {5 x^{2}-x y +y^{2}}{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 \(M=5 x^{2}-x y +y^{2}\) 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}-u +5\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} = 0 \]

Or

\[ -u \left (x \right )^{2}+u^{\prime }\left (x \right ) x +2 u \left (x \right )-5 = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}=\ln \left (x \right )+c_1 \]

Converting \(\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} -\frac {\arctan \left (\frac {-y+x}{2 x}\right )}{2} = \ln \left (x \right )+c_1 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Figure 2.64: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Summary of solutions found

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Solved as first order homogeneous class D2 ode

Time used: 0.106 (sec)

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {5 x^{2}-x^{2} u \left (x \right )+u \left (x \right )^{2} x^{2}}{x^{2}} \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}=\ln \left (x \right )+c_1 \]

Converting \(\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} -\frac {\arctan \left (-\frac {y}{2 x}+\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Figure 2.65: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Summary of solutions found

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Solved as first order homogeneous class Maple C ode

Time used: 0.363 (sec)

Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = \frac {5 \left (X +x_{0} \right )^{2}-\left (X +x_{0} \right ) \left (Y \left (X \right )+y_{0} \right )+\left (Y \left (X \right )+y_{0} \right )^{2}}{\left (X +x_{0} \right )^{2}} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=0\\ y_{0}&=0 \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = \frac {5 X^{2}-X Y \left (X \right )+Y \left (X \right )^{2}}{X^{2}} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= \frac {5 X^{2}-X Y +Y^{2}}{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 \(M=5 X^{2}-X Y +Y^{2}\) 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}-u +5\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X} \end{align*}

Or

\[ \frac {d}{d X}u \left (X \right )-\frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X} = 0 \]

Or

\[ \left (\frac {d}{d X}u \left (X \right )\right ) X -u \left (X \right )^{2}+2 u \left (X \right )-5 = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= \frac {1}{X}\\ g(u) &= u^{2}-2 u +5 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{X} \,dX} \\ \end{align*}

\[ \frac {\arctan \left (\frac {u \left (X \right )}{2}-\frac {1}{2}\right )}{2}=\ln \left (X \right )+c_1 \]

Converting \(\frac {\arctan \left (\frac {u \left (X \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (X \right )+c_1\) back to \(Y \left (X \right )\) gives

\begin{align*} -\frac {\arctan \left (\frac {-Y \left (X \right )+X}{2 X}\right )}{2} = \ln \left (X \right )+c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} -\frac {\arctan \left (\frac {-Y \left (X \right )+X}{2 X}\right )}{2} = \ln \left (X \right )+c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}

Or

\begin{align*} Y &= y\\ X &= x \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} -\frac {\arctan \left (\frac {-y+x}{2 x}\right )}{2} = \ln \left (x \right )+c_1 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Figure 2.66: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Solved as first order isobaric ode

Time used: 0.172 (sec)

Solving for \(y'\) gives

\begin{align*} \tag{1} y' &= \frac {5 x^{2}-x y+y^{2}}{x^{2}} \\ \end{align*}

Each of the above ode’s is now solved An ode \(y^{\prime }=f(x,y)\) is isobaric if

\[ f(t x, t^m y) = t^{m-1} f(x,y)\tag {1} \]

Where here

\[ f(x,y) = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = 1 \]

Since the ode is isobaric of order \(m=1\), then the substitution

\begin{align*} y&=u x^m \\ &=u x \end{align*}

Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives

\[ u \left (x \right )+x u^{\prime }\left (x \right ) = \frac {5 x^{2}-x^{2} u \left (x \right )+x^{2} u \left (x \right )^{2}}{x^{2}} \]

The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}=\ln \left (x \right )+c_1 \]

Converting \(\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} -\frac {\arctan \left (-\frac {y}{2 x}+\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Figure 2.67: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Summary of solutions found

\begin{align*} y &= 2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) x +x \\ \end{align*}

Solved using Lie symmetry for first order ode

Time used: 0.596 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\frac {5 x^{2}-x y +y^{2}}{x^{2}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+\frac {\left (5 x^{2}-x y +y^{2}\right ) \left (b_{3}-a_{2}\right )}{x^{2}}-\frac {\left (5 x^{2}-x y +y^{2}\right )^{2} a_{3}}{x^{4}}-\left (\frac {10 x -y}{x^{2}}-\frac {2 \left (5 x^{2}-x y +y^{2}\right )}{x^{3}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (-x +2 y \right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{x^{2}} = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {5 x^{4} a_{2}+25 x^{4} a_{3}-2 b_{2} x^{4}-5 x^{4} b_{3}-10 x^{3} y a_{3}+2 x^{3} y b_{2}-x^{2} y^{2} a_{2}+12 x^{2} y^{2} a_{3}+x^{2} y^{2} b_{3}-4 x \,y^{3} a_{3}+y^{4} a_{3}-x^{3} b_{1}+x^{2} y a_{1}+2 x^{2} y b_{1}-2 x \,y^{2} a_{1}}{x^{4}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -5 x^{4} a_{2}-25 x^{4} a_{3}+2 b_{2} x^{4}+5 x^{4} b_{3}+10 x^{3} y a_{3}-2 x^{3} y b_{2}+x^{2} y^{2} a_{2}-12 x^{2} y^{2} a_{3}-x^{2} y^{2} b_{3}+4 x \,y^{3} a_{3}-y^{4} a_{3}+x^{3} b_{1}-x^{2} y a_{1}-2 x^{2} y b_{1}+2 x \,y^{2} a_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -5 a_{2} v_{1}^{4}+a_{2} v_{1}^{2} v_{2}^{2}-25 a_{3} v_{1}^{4}+10 a_{3} v_{1}^{3} v_{2}-12 a_{3} v_{1}^{2} v_{2}^{2}+4 a_{3} v_{1} v_{2}^{3}-a_{3} v_{2}^{4}+2 b_{2} v_{1}^{4}-2 b_{2} v_{1}^{3} v_{2}+5 b_{3} v_{1}^{4}-b_{3} v_{1}^{2} v_{2}^{2}-a_{1} v_{1}^{2} v_{2}+2 a_{1} v_{1} v_{2}^{2}+b_{1} v_{1}^{3}-2 b_{1} v_{1}^{2} v_{2} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} \left (-5 a_{2}-25 a_{3}+2 b_{2}+5 b_{3}\right ) v_{1}^{4}+\left (10 a_{3}-2 b_{2}\right ) v_{1}^{3} v_{2}+b_{1} v_{1}^{3}+\left (a_{2}-12 a_{3}-b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-a_{1}-2 b_{1}\right ) v_{1}^{2} v_{2}+4 a_{3} v_{1} v_{2}^{3}+2 a_{1} v_{1} v_{2}^{2}-a_{3} v_{2}^{4} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} b_{1}&=0\\ 2 a_{1}&=0\\ -a_{3}&=0\\ 4 a_{3}&=0\\ -a_{1}-2 b_{1}&=0\\ 10 a_{3}-2 b_{2}&=0\\ a_{2}-12 a_{3}-b_{3}&=0\\ -5 a_{2}-25 a_{3}+2 b_{2}+5 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= x \\ \eta &= y \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y - \left (\frac {5 x^{2}-x y +y^{2}}{x^{2}}\right ) \left (x\right ) \\ &= \frac {-5 x^{2}+2 x y -y^{2}}{x}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {-5 x^{2}+2 x y -y^{2}}{x}}} dy \end{align*}

Which results in

\begin{align*} S&= -\frac {\arctan \left (\frac {2 y -2 x}{4 x}\right )}{2} \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= \frac {5 x^{2}-x y +y^{2}}{x^{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {y}{5 x^{2}-2 x y +y^{2}}\\ S_{y} &= -\frac {x}{5 x^{2}-2 x y +y^{2}} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {1}{x}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {1}{R} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {1}{R}\, dR}\\ S \left (R \right ) &= -\ln \left (R \right ) + c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {\arctan \left (\frac {-y+x}{2 x}\right )}{2} = -\ln \left (x \right )+c_2 \end{align*}

Which gives

\begin{align*} y = -2 \tan \left (-2 \ln \left (x \right )+2 c_2 \right ) x +x \end{align*}

The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.

Original ode in \(x,y\) coordinates	Canonical coordinates transformation	ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = \frac {5 x^{2}-x y +y^{2}}{x^{2}}\)		\( \frac {d S}{d R} = -\frac {1}{R}\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {\arctan \left (\frac {-y +x}{2 x}\right )}{2} \end {aligned} \)

Figure 2.68: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Summary of solutions found

\begin{align*} y &= -2 \tan \left (-2 \ln \left (x \right )+2 c_2 \right ) x +x \\ \end{align*}

Solved as first order ode of type Riccati

Time used: 0.141 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {5 x^{2}-x y +y^{2}}{x^{2}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = 5-\frac {y}{x}+\frac {y^{2}}{x^{2}} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=5\), \(f_1(x)=-\frac {1}{x}\) and \(f_2(x)=\frac {1}{x^{2}}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2}}} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=-\frac {2}{x^{3}}\\ f_1 f_2 &=-\frac {1}{x^{3}}\\ f_2^2 f_0 &=\frac {5}{x^{4}} \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2}}+\frac {3 u^{\prime }\left (x \right )}{x^{3}}+\frac {5 u \left (x \right )}{x^{4}} = 0 \end{align*}

This is Euler second order ODE. Let the solution be \(u = x^r\), then \(u'=r x^{r-1}\) and \(u''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives

\[ x^{2}(r(r-1))x^{r-2}+3 x r x^{r-1}+5 x^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )x^{r}+3 r\,x^{r}+5 x^{r} = 0 \]

Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives

\[ r \left (r -1\right )+3 r+5 = 0 \]

Or

\[ r^{2}+2 r +5 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= -1-2 i\\ r_2 &= -1+2 i \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha =-1\) and \(\beta =-2\). Hence the solution becomes

\begin{align*} u &= c_1 x^{r_1} + c_2 x^{r_2} \\ &= c_1 x^{\alpha + i \beta } + c_2 x^{\alpha - i \beta } \\ &= x^{\alpha } \left ( c_1 x^{i \beta } + c_2 x^{- i \beta }\right ) \\ &= x^{\alpha } \left ( c_1 e^{\ln \left (x^{i \beta }\right )} + c_2 e^{\ln \left (x^{-i \beta }\right )}\right ) \\ &= x^{\alpha } \left ( c_1 e^{i \left (\beta \ln {x}\right )} + c_2 e^{-i \left (\beta \ln {x}\right )}\right ) \end{align*}

Using the values for \(\alpha =-1, \beta =-2\), the above becomes

\begin{align*} u&= x^{-1} \left ( c_1 e^{-2 i \ln \left (x \right )} + c_2 e^{2 i \ln \left (x \right )} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} u&=\frac {1}{x}\left (c_1 \cos \left (2 \ln \left (x \right )\right )+c_2 \sin \left (2 \ln \left (x \right )\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = -\frac {c_1 \cos \left (2 \ln \left (x \right )\right )+c_2 \sin \left (2 \ln \left (x \right )\right )}{x^{2}}+\frac {-\frac {2 c_1 \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {2 c_2 \cos \left (2 \ln \left (x \right )\right )}{x}}{x} \]

Doing change of constants, the solution becomes

\[ y = -\frac {\left (-\frac {c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )}{x^{2}}+\frac {-\frac {2 c_3 \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {2 \cos \left (2 \ln \left (x \right )\right )}{x}}{x}\right ) x^{3}}{c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )} \]

Figure 2.69: Slope field \(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\)

Summary of solutions found

\begin{align*} y &= -\frac {\left (-\frac {c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )}{x^{2}}+\frac {-\frac {2 c_3 \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {2 \cos \left (2 \ln \left (x \right )\right )}{x}}{x}\right ) x^{3}}{c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )} \\ \end{align*}

✓ Maple. Time used: 0.006 (sec). Leaf size: 19

ode:=diff(y(x),x) = (5*x^2-x*y(x)+y(x)^2)/x^2; 
dsolve(ode,y(x), singsol=all);
 

\[ y = x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_{1} \right )\right ) \]

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`
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {5 x^{2}-x y \left (x \right )+y \left (x \right )^{2}}{x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {5 x^{2}-x y \left (x \right )+y \left (x \right )^{2}}{x^{2}} \end {array} \]

✓ Mathematica. Time used: 0.899 (sec). Leaf size: 18

ode=D[y[x],x]==(5*x^2-x*y[x]+y[x]^2)/x^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to x+2 x \tan (2 (\log (x)+c_1)) \]

✓ Sympy. Time used: 0.284 (sec). Leaf size: 34

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x) - (5*x**2 - x*y(x) + y(x)**2)/x**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = \frac {x \left (C_{1} \left (1 + 2 i\right ) + \left (-1 + 2 i\right ) e^{4 i \log {\left (x \right )}}\right )}{C_{1} - e^{4 i \log {\left (x \right )}}} \]