2.80 problem 656
Internal
problem
ID
[9640]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
Additional
non-linear
first
order
Problem
number
:
656
Date
solved
:
Friday, October 11, 2024 at 11:20:02 AM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
Solve
\begin{align*} y^{\prime }&=\frac {\left (\ln \left (y\right )+x^{3}\right ) y}{x} \end{align*}
2.80.1 Solved as first order Exact ode
Time used: 0.261 (sec)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows
that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}y} &= \left (\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\\ N(x,y) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the
following condition is satisfied
\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\right )\\ &= \frac {-x^{3}-\ln \left (y \right )-1}{x} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\) , then the ODE is not exact . Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( -\frac {1}{x}-\frac {\ln \left (y \right )+x^{3}}{x}\right ) - \left (0 \right ) \right ) \\ &=\frac {-x^{3}-\ln \left (y \right )-1}{x} \end{align*}
Since \(A\) depends on \(y\) , it can not be used to obtain an integrating factor. We will now try a
second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {x}{\left (\ln \left (y \right )+x^{3}\right ) y}\left ( \left ( 0\right ) - \left (-\frac {1}{x}-\frac {\ln \left (y \right )+x^{3}}{x} \right ) \right ) \\ &=\frac {-x^{3}-\ln \left (y \right )-1}{\left (\ln \left (y \right )+x^{3}\right ) y} \end{align*}
Since \(B\) depends on \(x\) , it can not be used to obtain an integrating factor.We will now try a third
method to find an integrating factor. Let
\[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \]
\(R\) is now checked to see if it is a function of only \(t=xy\) .
Therefore
\begin{align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (0\right )-\left (-\frac {1}{x}-\frac {\ln \left (y \right )+x^{3}}{x}\right )} {x\left (-\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\right ) - y\left (1\right )} \\ &= -\frac {1}{x y} \end{align*}
Replacing all powers of terms \(xy\) by \(t\) gives
\[ R = -\frac {1}{t} \]
Since \(R\) depends on \(t\) only, then it can be used to find
an integrating factor. Let the integrating factor be \(\mu \) then
\begin{align*} \mu &= e^{\int R \mathop {\mathrm {d}t}} \\ &= e^{\int \left (-\frac {1}{t}\right )\mathop {\mathrm {d}t} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-\ln \left (t \right ) } \\ &= \frac {1}{t} \end{align*}
Now \(t\) is replaced back with \(xy\) giving
\[ \mu =\frac {1}{x y} \]
Multiplying \(M\) and \(N\) by this integrating factor gives new \(M\)
and new \(N\) which are called \( \overline {M}\) and \( \overline {N}\) so not to confuse them with the original \(M\) and \(N\)
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{x y}\left (-\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\right ) \\ &= \frac {-\ln \left (y \right )-x^{3}}{x^{2}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{x y}\left (1\right ) \\ &= \frac {1}{x y} \end{align*}
A modified ODE is now obtained from the original ODE, which is exact and can solved. The
modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (\frac {-\ln \left (y \right )-x^{3}}{x^{2}}\right ) + \left (\frac {1}{x y}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {N}\tag {2} \end{align*}
Integrating (1) w.r.t. \(x\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \overline {M}\mathop {\mathrm {d}x} \\
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \frac {-\ln \left (y \right )-x^{3}}{x^{2}}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -\frac {x^{2}}{2}+\frac {\ln \left (y \right )}{x}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\) . Taking derivative of equation (3) w.r.t \(y\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = \frac {1}{x y}+f'(y)
\end{equation}
But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {1}{x y}\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {1}{x y} = \frac {1}{x y}+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[ f'(y) = 0 \]
Therefore
\[ f(y) = c_1 \]
Where \(c_1\) is
constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = -\frac {x^{2}}{2}+\frac {\ln \left (y \right )}{x}+ c_1
\]
But
since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining
\(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = -\frac {x^{2}}{2}+\frac {\ln \left (y \right )}{x}
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
\begin{align*} y = {\mathrm e}^{\frac {1}{2} x^{3}+c_1 x} \end{align*}
Figure 691: Slope field plot
\(y^{\prime } = \frac {\left (\ln \left (y\right )+x^{3}\right ) y}{x}\)
2.80.2 Solved using Lie symmetry for first order ode
Time used: 0.683 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\\ 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 (\ln \left (y \right )+x^{3}\right ) y \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{x}-\frac {\left (\ln \left (y \right )+x^{3}\right )^{2} y^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{x^{2}}-\left (3 y x -\frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x^{2}}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {1}{x}+\frac {\ln \left (y \right )+x^{3}}{x}\right ) \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {x^{7} y^{2} a_{5}+2 x^{6} y^{3} a_{6}+4 x^{5} y a_{4}+3 x^{4} y^{2} a_{5}-x^{4} y^{2} b_{6}+2 x^{3} y^{3} a_{6}+x \,y^{2} b_{6}+2 \ln \left (y \right )^{2} y^{3} a_{6}+\ln \left (y \right ) x^{3} b_{4}-\ln \left (y \right ) y^{3} a_{6}+x^{6} y^{2} a_{3}+3 x^{4} y a_{2}+2 x^{3} y^{2} a_{3}+2 x^{3} y a_{1}+x y b_{3}+\ln \left (y \right )^{2} y^{2} a_{3}+\ln \left (y \right ) x^{2} b_{2}-\ln \left (y \right ) y^{2} a_{3}+\ln \left (y \right ) x b_{1}-\ln \left (y \right ) y a_{1}+x^{5} b_{2}+x^{4} b_{1}+x b_{1}+2 \ln \left (y \right ) x^{4} y^{2} a_{5}+4 \ln \left (y \right ) x^{3} y^{3} a_{6}+\ln \left (y \right )^{2} x \,y^{2} a_{5}+\ln \left (y \right ) x^{2} y a_{4}-\ln \left (y \right ) x \,y^{2} b_{6}+2 \ln \left (y \right ) x^{3} y^{2} a_{3}-x^{3} b_{4}+x^{6} b_{4}}{x^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -x^{7} y^{2} a_{5}-2 x^{6} y^{3} a_{6}-4 x^{5} y a_{4}-3 x^{4} y^{2} a_{5}+x^{4} y^{2} b_{6}-2 x^{3} y^{3} a_{6}-x \,y^{2} b_{6}-2 \ln \left (y \right )^{2} y^{3} a_{6}-\ln \left (y \right ) x^{3} b_{4}+\ln \left (y \right ) y^{3} a_{6}-x^{6} y^{2} a_{3}-3 x^{4} y a_{2}-2 x^{3} y^{2} a_{3}-2 x^{3} y a_{1}-x y b_{3}-\ln \left (y \right )^{2} y^{2} a_{3}-\ln \left (y \right ) x^{2} b_{2}+\ln \left (y \right ) y^{2} a_{3}-\ln \left (y \right ) x b_{1}+\ln \left (y \right ) y a_{1}-x^{5} b_{2}-x^{4} b_{1}-x b_{1}-2 \ln \left (y \right ) x^{4} y^{2} a_{5}-4 \ln \left (y \right ) x^{3} y^{3} a_{6}-\ln \left (y \right )^{2} x \,y^{2} a_{5}-\ln \left (y \right ) x^{2} y a_{4}+\ln \left (y \right ) x \,y^{2} b_{6}-2 \ln \left (y \right ) x^{3} y^{2} a_{3}+x^{3} b_{4}-x^{6} b_{4} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, \ln \left (y \right )\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, \ln \left (y \right ) = v_{3}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -v_{1}^{7} v_{2}^{2} a_{5}-2 v_{1}^{6} v_{2}^{3} a_{6}-v_{1}^{6} v_{2}^{2} a_{3}-2 v_{3} v_{1}^{4} v_{2}^{2} a_{5}-4 v_{3} v_{1}^{3} v_{2}^{3} a_{6}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{3}-4 v_{1}^{5} v_{2} a_{4}-3 v_{1}^{4} v_{2}^{2} a_{5}-2 v_{1}^{3} v_{2}^{3} a_{6}-v_{1}^{6} b_{4}+v_{1}^{4} v_{2}^{2} b_{6}-3 v_{1}^{4} v_{2} a_{2}-2 v_{1}^{3} v_{2}^{2} a_{3}-v_{3}^{2} v_{1} v_{2}^{2} a_{5}-2 v_{3}^{2} v_{2}^{3} a_{6}-v_{1}^{5} b_{2}-2 v_{1}^{3} v_{2} a_{1}-v_{3}^{2} v_{2}^{2} a_{3}-v_{3} v_{1}^{2} v_{2} a_{4}+v_{3} v_{2}^{3} a_{6}-v_{1}^{4} b_{1}-v_{3} v_{1}^{3} b_{4}+v_{3} v_{1} v_{2}^{2} b_{6}+v_{3} v_{2}^{2} a_{3}-v_{3} v_{1}^{2} b_{2}+v_{1}^{3} b_{4}-v_{1} v_{2}^{2} b_{6}+v_{3} v_{2} a_{1}-v_{3} v_{1} b_{1}-v_{1} v_{2} b_{3}-v_{1} b_{1} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} -2 v_{3} v_{1}^{4} v_{2}^{2} a_{5}-4 v_{3} v_{1}^{3} v_{2}^{3} a_{6}-v_{3}^{2} v_{1} v_{2}^{2} a_{5}-v_{3} v_{1}^{2} v_{2} a_{4}+v_{3} v_{1} v_{2}^{2} b_{6}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{3}+\left (-3 a_{5}+b_{6}\right ) v_{1}^{4} v_{2}^{2}-v_{1}^{7} v_{2}^{2} a_{5}-2 v_{1}^{6} v_{2}^{3} a_{6}-4 v_{1}^{5} v_{2} a_{4}-2 v_{1}^{3} v_{2}^{3} a_{6}-v_{1} v_{2}^{2} b_{6}-2 v_{3}^{2} v_{2}^{3} a_{6}-v_{3} v_{1}^{3} b_{4}+v_{3} v_{2}^{3} a_{6}-v_{1}^{6} v_{2}^{2} a_{3}-3 v_{1}^{4} v_{2} a_{2}-2 v_{1}^{3} v_{2}^{2} a_{3}-2 v_{1}^{3} v_{2} a_{1}-v_{1} v_{2} b_{3}-v_{3}^{2} v_{2}^{2} a_{3}-v_{3} v_{1}^{2} b_{2}+v_{3} v_{2}^{2} a_{3}-v_{3} v_{1} b_{1}+v_{3} v_{2} a_{1}-v_{1}^{5} b_{2}-v_{1}^{4} b_{1}-v_{1} b_{1}+v_{1}^{3} b_{4}-v_{1}^{6} b_{4} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} a_{1}&=0\\ a_{3}&=0\\ a_{6}&=0\\ b_{4}&=0\\ b_{6}&=0\\ -2 a_{1}&=0\\ -3 a_{2}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ -4 a_{4}&=0\\ -a_{4}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ -b_{1}&=0\\ -b_{2}&=0\\ -b_{3}&=0\\ -b_{4}&=0\\ -b_{6}&=0\\ -3 a_{5}+b_{6}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=b_{5}\\ b_{6}&=0 \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 x \\
\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 x}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (y \right )}{x} \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 (\ln \left (y \right )+x^{3}\right ) y}{x} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {\ln \left (y \right )}{x^{2}}\\ S_{y} &= \frac {1}{x y} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= 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} &= 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 {R\, dR}\\ S \left (R \right ) &= \frac {R^{2}}{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 {\ln \left (y\right )}{x} = \frac {x^{2}}{2}+c_2 \end{align*}
Which gives
\begin{align*} y = {\mathrm e}^{\frac {1}{2} x^{3}+c_2 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 \((R,S)\)
\( \frac {dy}{dx} = \frac {\left (\ln \left (y \right )+x^{3}\right ) y}{x}\)
\( \frac {d S}{d R} = R\)
\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (y \right )}{x} \end {aligned} \)
Figure 692: Slope field plot
\(y^{\prime } = \frac {\left (\ln \left (y\right )+x^{3}\right ) y}{x}\)
2.80.3 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\left (\ln \left (y \left (x \right )\right )+x^{3}\right ) y \left (x \right )}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\left (\ln \left (y \left (x \right )\right )+x^{3}\right ) y \left (x \right )}{x} \end {array} \]
2.80.4 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 Chini
differential order: 1; looking for linear symmetries
trying exact
Looking for potential symmetries
trying inverse_Riccati
trying an equivalence to an Abel ODE
differential order: 1; trying a linearization to 2nd order
--- trying a change of variables {x -> y(x), y(x) -> x}
differential order: 1; trying a linearization to 2nd order
trying 1st order ODE linearizable_by_differentiation
--- Trying Lie symmetry methods, 1st order ---
` , ` -> Computing symmetries using: way = 3 ` [0, y*x]
2.80.5 Maple dsolve solution
Solving time : 0.007
(sec)
Leaf size : 15
dsolve ( diff ( y ( x ), x ) = ( ln ( y ( x ))+ x ^3)* y ( x )/ x ,
y(x),singsol=all)
\[
y = {\mathrm e}^{\frac {x \left (x^{2}+2 c_{1} \right )}{2}}
\]
2.80.6 Mathematica DSolve solution
Solving time : 0.24
(sec)
Leaf size : 20
DSolve [{ D [ y [ x ], x ] == ((x^3 + Log [y[x]])*y[x])/x,{}},
y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to e^{\frac {x^3}{2}+3 c_1 x}
\]