Internal problem ID [9497]
Internal file name [OUTPUT/8437_Monday_June_06_2022_03_04_23_AM_24943982/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1168.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_euler_ode", "exact linear second order ode", "second_order_integrable_as_is", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x^{2} y^{\prime \prime }+2 y^{\prime } x=0} \] The ODE is \[ x^{2} y^{\prime \prime }+2 y^{\prime } x = 0 \] Or \[ x \left (y^{\prime \prime } x +2 y^{\prime }\right ) = 0 \] For \(x \neq 0\) the above simplifies to \[ y^{\prime \prime } x +2 y^{\prime } = 0 \]
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 \[ x^{2}(r(r-1))x^{r-2}+2 x r x^{r-1}+0 = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}+2 r\,x^{r}+0 = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )+2 r+0 = 0 \] Or \[ r^{2}+r = 0 \tag {1} \] 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 &= 0\\ r_2 &= -1 \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 \[ y = c_{1} +\frac {c_{2}}{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +\frac {c_{2}}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} +\frac {c_{2}}{x} \] Verified OK.
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x^{2} y^{\prime \prime }+2 y^{\prime } x \right )d x &= 0 \\ x^{2} y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{1}}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{x}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1}}{x}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {c_{1}}{x}+c_{2} \] Verified OK.
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} x^{2} p^{\prime }\left (x \right )+2 p \left (x \right ) x = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {2 p}{x} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {2}{x} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {2}{x} \,d x}\\ \ln \left (p \right )&=-2 \ln \left (x \right )+c_{1}\\ p&={\mathrm e}^{-2 \ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x^{2}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {c_{1}}{x^{2}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {c_{1}}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{x}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1}}{x}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {c_{1}}{x}+c_{2} \] Verified OK.
Writing the ode as \[ x^{2} y^{\prime \prime }+2 y^{\prime } x = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x^{2} y^{\prime \prime }+2 y^{\prime } x \right )d x &= 0 \\ x^{2} y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{1}}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{x}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1}}{x}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {c_{1}}{x}+c_{2} \] Verified OK.
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+2 y^{\prime } x &= 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 &= 2 x\tag {3} \\ C &= 0 \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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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 {2 x}{x^{2}} \,dx} \\ &= z_1 e^{-\ln \left (x \right )} \\ &= z_1 \left (\frac {1}{x}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {1}{x} \] 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 {2 x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-2 \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 (\frac {1}{x}\right ) + c_{2} \left (\frac {1}{x}\left (x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x}+c_{2} \] Verified OK.
An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}
is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}
For the given ode we have \begin {align*} p(x) &= x^{2}\\ q(x) &= 2 x\\ r(x) &= 0\\ s(x) &= 0 \end {align*}
Hence \begin {align*} p''(x) &= 2\\ q'(x) &= 2 \end {align*}
Therefore (1) becomes \begin {align*} 2- \left (2\right ) + \left (0\right )&=0 \end {align*}
Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}
Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}
Substituting the above values for \(p,q,r,s\) gives \begin {align*} x^{2} y^{\prime }&=c_{1} \end {align*}
We now have a first order ode to solve which is \begin {align*} x^{2} y^{\prime } = c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {c_{1}}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{x}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1}}{x}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {c_{1}}{x}+c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+2 y^{\prime } x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {2 y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {2 y^{\prime }}{x}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x +2 y^{\prime }=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} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\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} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+t^{\prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\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} \\ {} & {} & \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 ) x +\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x}=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\frac {d^{2}}{d t^{2}}y \left (t \right )+\frac {d}{d t}y \left (t \right )}{x}=0 \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=-\frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \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 t^{2}}y \left (t \right )+\frac {d}{d t}y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 0\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=1 \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} {\mathrm e}^{-t}+c_{2} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {c_{1}}{x}+c_{2} \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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 11
dsolve(x^2*diff(diff(y(x),x),x)+2*x*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +\frac {c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 15
DSolve[2*x*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2-\frac {c_1}{x} \]