2.2 problem Ex 2

Internal problem ID [11772]
Book : An elementary treatise on differential equations by Abraham Cohen. DC heath publishers. 1906
Section : Chapter 2, differential equations of the first order and the first degree. Article 9. Variables searated or separable. Page 13
Problem number : Ex 2
Date solved : Friday, October 18, 2024 at 07:24:00 AM
CAS classification : [_separable]

Solve

\begin{align*} \left (1+x \right ) y^{2}-x^{3} y^{\prime }&=0 \end{align*}

2.2.1 Solved as first order separable ode

Time used: 0.094 (sec)

The ode \(y^{\prime } = \frac {\left (1+x \right ) y^{2}}{x^{3}}\) is separable as it can be written as

\begin{align*} y^{\prime }&= \frac {\left (1+x \right ) y^{2}}{x^{3}}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= \frac {1+x}{x^{3}}\\ g(y) &= y^{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{y^{2}}\,dy} &= \int { \frac {1+x}{x^{3}} \,dx}\\ -\frac {1}{y}&=\frac {-2 x -1}{2 x^{2}}+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or \(y^{2}=0\) for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {1}{y} = \frac {-2 x -1}{2 x^{2}}+c_1\\ y = 0 \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=0\\ y&=-\frac {2 x^{2}}{2 c_1 \,x^{2}-2 x -1} \end{align*}

Figure 28: Slope field plot
\(\left (1+x \right ) y^{2}-x^{3} y^{\prime } = 0\)

2.2.2 Solved as first order Bernoulli ode

Time used: 0.066 (sec)

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {\left (1+x \right ) y^{2}}{x^{3}} \end{align*}

This is a Bernoulli ODE.

\[ y' = \left (\frac {1+x}{x^{3}}\right ) y^{2} \tag {1} \]

The standard Bernoulli ODE has the form

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

Comparing this to (1) shows that

\begin{align*} f_0 &=0\\ f_1 &=\frac {1+x}{x^{3}} \end{align*}

The first step is to divide the above equation by \(y^n \) which gives

\[ \frac {y'}{y^n} = f_1(x) \tag {3} \]

The next step is use the substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=0\\ f_1(x)&=\frac {1+x}{x^{3}}\\ n &=2 \end{align*}

Dividing both sides of ODE (1) by \(y^n=y^{2}\) gives

\begin{align*} y'\frac {1}{y^{2}} &= 0 +\frac {1+x}{x^{3}} \tag {4} \end{align*}

Let

\begin{align*} v &= y^{1-n} \\ &= \frac {1}{y} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= -\frac {1}{y^{2}}y' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -v^{\prime }\left (x \right )&= \frac {1+x}{x^{3}}\\ v' &= -\frac {1+x}{x^{3}} \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

Since the ode has the form \(v^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dv} &= \int {-\frac {1+x}{x^{3}}\, dx}\\ v \left (x \right ) &= \frac {1}{x}+\frac {1}{2 x^{2}} + c_1 \end{align*}

The substitution \(v = y^{1-n}\) is now used to convert the above solution back to \(y\) which results in

\[ \frac {1}{y} = \frac {1}{x}+\frac {1}{2 x^{2}}+c_1 \]

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y = \frac {2 x^{2}}{2 c_1 \,x^{2}+2 x +1} \end{align*}

Figure 29: Slope field plot
\(\left (1+x \right ) y^{2}-x^{3} y^{\prime } = 0\)

2.2.3 Solved using Lie symmetry for first order ode

Time used: 0.664 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\frac {\left (1+x \right ) y^{2}}{x^{3}}\\ 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 2 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \]

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

\begin{equation} \tag{5E} 2 x b_{4}+y b_{5}+b_{2}+\frac {\left (1+x \right ) y^{2} \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{x^{3}}-\frac {\left (1+x \right )^{2} y^{4} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{x^{6}}-\left (\frac {y^{2}}{x^{3}}-\frac {3 \left (1+x \right ) y^{2}}{x^{4}}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\frac {2 \left (1+x \right ) y \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{x^{3}} = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {2 x^{7} b_{4}-2 x^{6} y b_{4}+y b_{5} x^{6}-x^{5} y^{2} b_{5}+x^{4} y^{3} a_{5}-x^{3} y^{4} a_{5}+2 x^{3} y^{4} a_{6}-2 x^{2} y^{5} a_{6}+b_{2} x^{6}-2 x^{5} y b_{2}-2 x^{5} y b_{4}+x^{4} y^{2} a_{2}+x^{4} y^{2} a_{4}-x^{4} y^{2} b_{3}-x^{4} y^{2} b_{5}+2 x^{3} y^{3} a_{3}+2 x^{3} y^{3} a_{5}-x^{2} y^{4} a_{3}-2 x^{2} y^{4} a_{5}+3 x^{2} y^{4} a_{6}-4 x \,y^{5} a_{6}-2 x^{4} y b_{1}-2 x^{4} y b_{2}+2 x^{3} y^{2} a_{1}+2 x^{3} y^{2} a_{2}-x^{3} y^{2} b_{3}+3 x^{2} y^{3} a_{3}-2 x \,y^{4} a_{3}-x \,y^{4} a_{5}-2 y^{5} a_{6}-2 x^{3} y b_{1}+3 x^{2} y^{2} a_{1}-y^{4} a_{3}}{x^{6}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 2 x^{7} b_{4}-2 x^{6} y b_{4}+y b_{5} x^{6}-x^{5} y^{2} b_{5}+x^{4} y^{3} a_{5}-x^{3} y^{4} a_{5}+2 x^{3} y^{4} a_{6}-2 x^{2} y^{5} a_{6}+b_{2} x^{6}-2 x^{5} y b_{2}-2 x^{5} y b_{4}+x^{4} y^{2} a_{2}+x^{4} y^{2} a_{4}-x^{4} y^{2} b_{3}-x^{4} y^{2} b_{5}+2 x^{3} y^{3} a_{3}+2 x^{3} y^{3} a_{5}-x^{2} y^{4} a_{3}-2 x^{2} y^{4} a_{5}+3 x^{2} y^{4} a_{6}-4 x \,y^{5} a_{6}-2 x^{4} y b_{1}-2 x^{4} y b_{2}+2 x^{3} y^{2} a_{1}+2 x^{3} y^{2} a_{2}-x^{3} y^{2} b_{3}+3 x^{2} y^{3} a_{3}-2 x \,y^{4} a_{3}-x \,y^{4} a_{5}-2 y^{5} a_{6}-2 x^{3} y b_{1}+3 x^{2} y^{2} a_{1}-y^{4} a_{3} = 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} a_{5} v_{1}^{4} v_{2}^{3}-a_{5} v_{1}^{3} v_{2}^{4}+2 a_{6} v_{1}^{3} v_{2}^{4}-2 a_{6} v_{1}^{2} v_{2}^{5}+2 b_{4} v_{1}^{7}-2 b_{4} v_{1}^{6} v_{2}+b_{5} v_{1}^{6} v_{2}-b_{5} v_{1}^{5} v_{2}^{2}+a_{2} v_{1}^{4} v_{2}^{2}+2 a_{3} v_{1}^{3} v_{2}^{3}-a_{3} v_{1}^{2} v_{2}^{4}+a_{4} v_{1}^{4} v_{2}^{2}+2 a_{5} v_{1}^{3} v_{2}^{3}-2 a_{5} v_{1}^{2} v_{2}^{4}+3 a_{6} v_{1}^{2} v_{2}^{4}-4 a_{6} v_{1} v_{2}^{5}+b_{2} v_{1}^{6}-2 b_{2} v_{1}^{5} v_{2}-b_{3} v_{1}^{4} v_{2}^{2}-2 b_{4} v_{1}^{5} v_{2}-b_{5} v_{1}^{4} v_{2}^{2}+2 a_{1} v_{1}^{3} v_{2}^{2}+2 a_{2} v_{1}^{3} v_{2}^{2}+3 a_{3} v_{1}^{2} v_{2}^{3}-2 a_{3} v_{1} v_{2}^{4}-a_{5} v_{1} v_{2}^{4}-2 a_{6} v_{2}^{5}-2 b_{1} v_{1}^{4} v_{2}-2 b_{2} v_{1}^{4} v_{2}-b_{3} v_{1}^{3} v_{2}^{2}+3 a_{1} v_{1}^{2} v_{2}^{2}-a_{3} v_{2}^{4}-2 b_{1} v_{1}^{3} 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} 2 b_{4} v_{1}^{7}+\left (-2 b_{4}+b_{5}\right ) v_{1}^{6} v_{2}+b_{2} v_{1}^{6}-b_{5} v_{1}^{5} v_{2}^{2}+\left (-2 b_{2}-2 b_{4}\right ) v_{1}^{5} v_{2}+a_{5} v_{1}^{4} v_{2}^{3}+\left (a_{2}+a_{4}-b_{3}-b_{5}\right ) v_{1}^{4} v_{2}^{2}+\left (-2 b_{1}-2 b_{2}\right ) v_{1}^{4} v_{2}+\left (-a_{5}+2 a_{6}\right ) v_{1}^{3} v_{2}^{4}+\left (2 a_{3}+2 a_{5}\right ) v_{1}^{3} v_{2}^{3}+\left (2 a_{1}+2 a_{2}-b_{3}\right ) v_{1}^{3} v_{2}^{2}-2 b_{1} v_{1}^{3} v_{2}-2 a_{6} v_{1}^{2} v_{2}^{5}+\left (-a_{3}-2 a_{5}+3 a_{6}\right ) v_{1}^{2} v_{2}^{4}+3 a_{3} v_{1}^{2} v_{2}^{3}+3 a_{1} v_{1}^{2} v_{2}^{2}-4 a_{6} v_{1} v_{2}^{5}+\left (-2 a_{3}-a_{5}\right ) v_{1} v_{2}^{4}-2 a_{6} v_{2}^{5}-a_{3} v_{2}^{4} = 0 \end{equation}

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

\begin{align*} a_{5}&=0\\ b_{2}&=0\\ 3 a_{1}&=0\\ -a_{3}&=0\\ 3 a_{3}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ -2 b_{1}&=0\\ 2 b_{4}&=0\\ -b_{5}&=0\\ -2 a_{3}-a_{5}&=0\\ 2 a_{3}+2 a_{5}&=0\\ -a_{5}+2 a_{6}&=0\\ -2 b_{1}-2 b_{2}&=0\\ -2 b_{2}-2 b_{4}&=0\\ -2 b_{4}+b_{5}&=0\\ 2 a_{1}+2 a_{2}-b_{3}&=0\\ -a_{3}-2 a_{5}+3 a_{6}&=0\\ a_{2}+a_{4}-b_{3}-b_{5}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=\frac {b_{3}}{2}\\ a_{3}&=0\\ a_{4}&=\frac {b_{3}}{2}\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3}\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=b_{6} \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 &= 0 \\ \eta &= y^{2} \\ \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}{y^{2}}} dy \end{align*}

Which results in

\begin{align*} S&= -\frac {1}{y} \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 {\left (1+x \right ) y^{2}}{x^{3}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= 0\\ S_{y} &= \frac {1}{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}{x^{3}}\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}{R^{3}} \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}{R^{3}}\, dR}\\ S \left (R \right ) &= -\frac {1}{R}-\frac {1}{2 R^{2}} + 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 {1}{y} = -\frac {1}{x}-\frac {1}{2 x^{2}}+c_2 \end{align*}

Which gives

\begin{align*} y = -\frac {2 x^{2}}{2 c_2 \,x^{2}-2 x -1} \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 \((R,S)\)
\( \frac {dy}{dx} = \frac {\left (1+x \right ) y^{2}}{x^{3}}\)		\( \frac {d S}{d R} = \frac {1+R}{R^{3}}\)
	\(\!\begin {aligned} R&= x\\ S&= -\frac {1}{y} \end {aligned} \)

Figure 30: Slope field plot
\(\left (1+x \right ) y^{2}-x^{3} y^{\prime } = 0\)

2.2.4 Solved as first order ode of type Riccati

Time used: 0.112 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {\left (1+x \right ) y^{2}}{x^{3}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \frac {y^{2}}{x^{3}}+\frac {y^{2}}{x^{2}} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=0\), \(f_1(x)=0\) and \(f_2(x)=\frac {1+x}{x^{3}}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u \left (1+x \right )}{x^{3}}} \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 {1}{x^{3}}-\frac {3 \left (1+x \right )}{x^{4}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=0 \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} \frac {\left (1+x \right ) u^{\prime \prime }\left (x \right )}{x^{3}}-\left (\frac {1}{x^{3}}-\frac {3 \left (1+x \right )}{x^{4}}\right ) u^{\prime }\left (x \right ) = 0 \end{align*}

This is second order ode with missing dependent variable \(u\). Let

\begin{align*} p(x) &= \frac {d u}{d x} \end{align*}

Then

\begin{align*} p'(x) &= \frac {d^{2}u}{d x^{2}} \end{align*}

Hence the ode becomes

\begin{align*} \frac {\left (1+x \right ) \left (\frac {d}{d x}p \left (x \right )\right )}{x^{3}}-\left (\frac {1}{x^{3}}-\frac {3 \left (1+x \right )}{x^{4}}\right ) p \left (x \right ) = 0 \end{align*}

Which is now solve for \(p(x)\) as first order ode. In canonical form a linear first order is

\begin{align*} \frac {d}{d x}p \left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {2 x +3}{x \left (1+x \right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {2 x +3}{x \left (1+x \right )}d x}\\ &= \frac {x^{3}}{1+x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {p \,x^{3}}{1+x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {p \,x^{3}}{1+x}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\frac {x^{3}}{1+x}\) gives the final solution

\[ p \left (x \right ) = \frac {\left (1+x \right ) c_1}{x^{3}} \]

For solution (1) found earlier, since \(p=\frac {d u}{d x}\) then we now have a new first order ode to solve which is

\begin{align*} \frac {d u}{d x} = \frac {\left (1+x \right ) c_1}{x^{3}} \end{align*}

Since the ode has the form \(\frac {d u}{d x}=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {du} &= \int {\frac {\left (1+x \right ) c_1}{x^{3}}\, dx}\\ u &= c_1 \left (-\frac {1}{x}-\frac {1}{2 x^{2}}\right ) + c_2 \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {4 c_2 x -2 c_1}{2 x^{2}}-\frac {2 c_2 \,x^{2}-2 c_1 x -c_1}{x^{3}} \]

Doing change of constants, the solution becomes

\[ y = -\frac {2 \left (\frac {-2 c_3 +4 x}{2 x^{2}}-\frac {-2 c_3 x +2 x^{2}-c_3}{x^{3}}\right ) x^{5}}{\left (1+x \right ) \left (-2 c_3 x +2 x^{2}-c_3 \right )} \]

Figure 31: Slope field plot
\(\left (1+x \right ) y^{2}-x^{3} y^{\prime } = 0\)

2.2.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +1\right ) y \left (x \right )^{2}-x^{3} \left (\frac {d}{d x}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\left (x +1\right ) y \left (x \right )^{2}}{x^{3}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )^{2}}=\frac {x +1}{x^{3}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )^{2}}d x =\int \frac {x +1}{x^{3}}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y \left (x \right )}=-\frac {1}{2 x^{2}}-\frac {1}{x}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {2 x^{2}}{2 \mathit {C1} \,x^{2}-2 x -1} \end {array} \]

2.2.6 Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

2.2.7 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 22

dsolve((x+1)*y(x)^2-diff(y(x),x)*x^3 = 0, 
       y(x),singsol=all)
 

\[ y = \frac {2 x^{2}}{2 c_1 \,x^{2}+2 x +1} \]

2.2.8 Mathematica DSolve solution

Solving time : 0.143 (sec)
Leaf size : 29

DSolve[{(1+x)*y[x]^2-x^3*D[y[x],x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \frac {2 x^2}{-2 c_1 x^2+2 x+1} \\ y(x)\to 0 \\ \end{align*}