2.2.46 problem 46

Internal problem ID [8581]
Book : Second order enumerated odes
Section : section 2
Problem number : 46
Date solved : Sunday, November 10, 2024 at 04:05:32 AM
CAS classification : [[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solve

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y&=0 \end{align*}

Solved as second order Euler type ode

Time used: 0.050 (sec)

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

\[ x^{2}(r(r-1))x^{r-2}-4 x r x^{r-1}+6 x^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )x^{r}-4 r\,x^{r}+6 x^{r} = 0 \]

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

\[ r \left (r -1\right )-4 r+6 = 0 \]

Or

\[ r^{2}-5 r +6 = 0 \tag {1} \]

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 &= 2\\ r_2 &= 3 \end{align*}

Since the roots are real and distinct, then the general solution is

\[ y= c_1 y_1 + c_2 y_2 \]

Where \(y_1 = x^{r_1}\) and \(y_2 = x^{r_2} \). Hence

\[ y = c_2 \,x^{3}+c_1 \,x^{2} \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_2 \,x^{3}+c_1 \,x^{2} \\ \end{align*}

Solved as second order solved by an integrating factor

Time used: 0.031 (sec)

The ode satisfies this form

\[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \]

Where \( p(x) = -\frac {4}{x}\). Therefore, there is an integrating factor given by

\begin{align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int -\frac {4}{x} \, dx} \\ &= \frac {1}{x^{2}} \end{align*}

Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential

\begin{align*} \left ( M(x) y \right )'' &= 0 \\ \left ( \frac {y}{x^{2}} \right )'' &= 0 \\ \end{align*}

Integrating once gives

\[ \left ( \frac {y}{x^{2}} \right )' = c_1 \]

Integrating again gives

\[ \left ( \frac {y}{x^{2}} \right ) = c_1 x +c_2 \]

Hence the solution is

\begin{align*} y &= \frac {c_1 x +c_2}{\frac {1}{x^{2}}} \\ \end{align*}

Or

\[ y = c_1 \,x^{3}+c_2 \,x^{2} \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \,x^{3}+c_2 \,x^{2} \\ \end{align*}

Solved as second order ode using change of variable on x method 2

Time used: 0.339 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {4}{x}\\ q \left (x \right )&=\frac {6}{x^{2}} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int -\frac {4}{x}d x \right )}d x\\ &= \int e^{4 \ln \left (x \right )} \,dx\\ &= \int x^{4}d x\\ &= \frac {x^{5}}{5}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {6}{x^{2}}}{x^{8}}\\ &= \frac {6}{x^{10}}\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {6 y \left (\tau \right )}{x^{10}}&=0 \\ \end{align*}

But in terms of \(\tau \)

\begin{align*} \frac {6}{x^{10}}&=\frac {6}{25 \tau ^{2}} \end{align*}

Hence the above ode becomes

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {6 y \left (\tau \right )}{25 \tau ^{2}}&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives

\[ 25 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+6 \tau ^{r} = 0 \]

Simplifying gives

\[ 25 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+6 \tau ^{r} = 0 \]

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

\[ 25 r \left (r -1\right )+0+6 = 0 \]

Or

\[ 25 r^{2}-25 r +6 = 0 \tag {1} \]

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 &= {\frac {2}{5}}\\ r_2 &= {\frac {3}{5}} \end{align*}

Since the roots are real and distinct, then the general solution is

\[ y \left (\tau \right )= c_1 y_1 + c_2 y_2 \]

Where \(y_1 = \tau ^{r_1}\) and \(y_2 = \tau ^{r_2} \). Hence

\[ y \left (\tau \right ) = c_1 \,\tau ^{{2}/{5}}+c_2 \,\tau ^{{3}/{5}} \]

Will add steps showing solving for IC soon.

The above solution is now transformed back to \(y\) using (6) which results in

\[ y = \frac {c_1 5^{{3}/{5}} \left (x^{5}\right )^{{2}/{5}}}{5}+\frac {c_2 5^{{2}/{5}} \left (x^{5}\right )^{{3}/{5}}}{5} \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {c_1 5^{{3}/{5}} \left (x^{5}\right )^{{2}/{5}}}{5}+\frac {c_2 5^{{2}/{5}} \left (x^{5}\right )^{{3}/{5}}}{5} \\ \end{align*}

Solved as second order ode using change of variable on x method 1

Time used: 0.124 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {4}{x}\\ q \left (x \right )&=\frac {6}{x^{2}} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) results

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5)

\begin{align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {6}\, \sqrt {\frac {1}{x^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {6}}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}} \end{align*}

Substituting the above into (4) results in

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {-\frac {\sqrt {6}}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}}-\frac {4}{x}\frac {\sqrt {6}\, \sqrt {\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {6}\, \sqrt {\frac {1}{x^{2}}}}{c}\right )^2} \\ &=-\frac {5 c \sqrt {6}}{6} \end{align*}

Therefore ode (3) now becomes

\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )-\frac {5 c \sqrt {6}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{6}+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give

\begin{align*} y \left (\tau \right ) &= {\mathrm e}^{\frac {5 \sqrt {6}\, c \tau }{12}} \left (c_1 \cosh \left (\frac {\sqrt {6}\, c \tau }{12}\right )+i c_2 \sinh \left (\frac {\sqrt {6}\, c \tau }{12}\right )\right ) \end{align*}

Now from (6)

\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {6}\, \sqrt {\frac {1}{x^{2}}}d x}{c}\\ &= \frac {\sqrt {6}\, \ln \left (x \right )}{c} \end{align*}

Substituting the above into the solution obtained gives

\[ y = x^{{5}/{2}} \left (c_1 \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_2 \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= x^{{5}/{2}} \left (c_1 \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_2 \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \\ \end{align*}

Solved as second order ode using change of variable on y method 1

Time used: 0.060 (sec)

In normal form the given ode is written as

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {4}{x}\\ q \left (x \right )&=\frac {6}{x^{2}} \end{align*}

Calculating the Liouville ode invariant \(Q\) given by

\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {6}{x^{2}} - \frac {\left (-\frac {4}{x}\right )'}{2}- \frac {\left (-\frac {4}{x}\right )^2}{4} \\ &= \frac {6}{x^{2}} - \frac {\left (\frac {4}{x^{2}}\right )}{2}- \frac {\left (\frac {16}{x^{2}}\right )}{4} \\ &= \frac {6}{x^{2}} - \left (\frac {2}{x^{2}}\right )-\frac {4}{x^{2}}\\ &= 0 \end{align*}

Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation

\begin{align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end{align*}

is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by

\begin{align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {-\frac {4}{x}}{2} }\\ &= x^{2}\tag {5} \end{align*}

Hence (3) becomes

\begin{align*} y = v \left (x \right ) x^{2}\tag {4} \end{align*}

Applying this change of variable to the original ode results in

\begin{align*} x^{4} v^{\prime \prime }\left (x \right ) = 0 \end{align*}

Which is now solved for \(v \left (x \right )\).

The above ode can be simplified to

\begin{align*} v^{\prime \prime }\left (x \right ) = 0 \end{align*}

Integrating twice gives the solution

\[ v \left (x \right )= c_1 x + c_2 \]

Will add steps showing solving for IC soon.

Now that \(v \left (x \right )\) is known, then

\begin{align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 x +c_2\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}

But from (5)

\begin{align*} z \left (x \right )&= x^{2} \end{align*}

Hence (7) becomes

\begin{align*} y = \left (c_1 x +c_2 \right ) x^{2} \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \left (c_1 x +c_2 \right ) x^{2} \\ \end{align*}

Solved as second order ode using change of variable on y method 2

Time used: 0.163 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {4}{x}\\ q \left (x \right )&=\frac {6}{x^{2}} \end{align*}

Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\).

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}-\frac {4 n}{x^{2}}+\frac {6}{x^{2}}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=3 \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\frac {2 v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {2 v^{\prime }\left (x \right )}{x}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (x \right )+\frac {2 u \left (x \right )}{x} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). In canonical form a linear first order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {2}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {2}{x}d x}\\ &= x^{2} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,x^{2}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,x^{2}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(x^{2}\) gives the final solution

\[ u \left (x \right ) = \frac {c_1}{x^{2}} \]

Now that \(u \left (x \right )\) is known, then

\begin{align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= -\frac {c_1}{x}+c_2 \end{align*}

Hence

\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_1}{x}+c_2 \right ) x^{3}\\ &= \left (c_2 x -c_1 \right ) x^{2}\\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \left (-\frac {c_1}{x}+c_2 \right ) x^{3} \\ \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.039 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= x^{2} \\ B &= -4 x\tag {3} \\ C &= 6 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 0\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= 0 \tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.53: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(x) = 1 \]

Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-4 x}{x^{2}} \,dx} \\ &= z_1 e^{2 \ln \left (x \right )} \\ &= z_1 \left (x^{2}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = x^{2} \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]

Substituting gives

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-4 x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{4 \ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (x^{2}\right ) + c_2 \left (x^{2}\left (x\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_2 \,x^{3}+c_1 \,x^{2} \\ \end{align*}

Solved as second order ode adjoint method

Time used: 0.122 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }-4 x y^{\prime }+6 y = 0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {4}{x}\\ q \left (x \right )&=\frac {6}{x^{2}}\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (-\frac {4 \xi \left (x \right )}{x}\right )' + \left (\frac {6 \xi \left (x \right )}{x^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )+\frac {2 \xi \left (x \right )}{x^{2}}+\frac {4 \xi ^{\prime }\left (x \right )}{x}&= 0 \end{align*}

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

\[ x^{2}(r(r-1))x^{r-2}+4 x r x^{r-1}+2 x^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )x^{r}+4 r\,x^{r}+2 x^{r} = 0 \]

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

\[ r \left (r -1\right )+4 r+2 = 0 \]

Or

\[ r^{2}+3 r +2 = 0 \tag {1} \]

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 &= -2\\ r_2 &= -1 \end{align*}

Since the roots are real and distinct, then the general solution is

\[ \xi = c_1 \xi _1 + c_2 \xi _2 \]

Where \(\xi _1 = x^{r_1}\) and \(\xi _2 = x^{r_2} \). Hence

\[ \xi = \frac {c_1}{x^{2}}+\frac {c_2}{x} \]

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (-\frac {4}{x}-\frac {-\frac {2 c_1}{x^{3}}-\frac {c_2}{x^{2}}}{\frac {c_1}{x^{2}}+\frac {c_2}{x}}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {3 c_2 x +2 c_1}{x \left (c_2 x +c_1 \right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {3 c_2 x +2 c_1}{x \left (c_2 x +c_1 \right )}d x}\\ &= \frac {1}{\left (c_2 x +c_1 \right ) x^{2}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{\left (c_2 x +c_1 \right ) x^{2}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {y}{\left (c_2 x +c_1 \right ) x^{2}}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{\left (c_2 x +c_1 \right ) x^{2}}\) gives the final solution

\[ y = \left (c_2 x +c_1 \right ) x^{2} c_3 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_2 x +c_1 \right ) x^{2} c_3 \\ \end{align*}

The constants can be merged to give

\[ y = x^{2} \left (c_2 x +c_1 \right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= x^{2} \left (c_2 x +c_1 \right ) \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-4 x \left (\frac {d}{d x}y \left (x \right )\right )+6 y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {6 y \left (x \right )}{x^{2}}+\frac {4 \left (\frac {d}{d x}y \left (x \right )\right )}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )-\frac {4 \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {6 y \left (x \right )}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-4 x \left (\frac {d}{d x}y \left (x \right )\right )+6 y \left (x \right )=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (\frac {d}{d t}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )^{2}+\left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )-4 \frac {d}{d t}y \left (t \right )+6 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )-5 \frac {d}{d t}y \left (t \right )+6 y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-5 r +6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right ) \left (r -3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2, 3\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{2 t}+\mathit {C2} \,{\mathrm e}^{3 t} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y \left (x \right )=\mathit {C2} \,x^{3}+\mathit {C1} \,x^{2} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=x^{2} \left (\mathit {C2} x +\mathit {C1} \right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
<- LODE of Euler type successful`
 

Maple dsolve solution

Solving time : 0.006 (sec)
Leaf size : 13

dsolve(x^2*diff(diff(y(x),x),x)-4*diff(y(x),x)*x+6*y(x) = 0, 
       y(x),singsol=all)
 

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

Mathematica DSolve solution

Solving time : 0.016 (sec)
Leaf size : 16

DSolve[{x^2*D[y[x],{x,2}]-4*x*D[y[x],x]+6*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to x^2 (c_2 x+c_1) \]