Internal problem ID [6006]
Internal file name [OUTPUT/5254_Sunday_June_05_2022_03_28_40_PM_67813326/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 3. Linear equations with variable coefficients. Page 108
Problem number: 1(c.1).
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_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {y^{\prime \prime }+\frac {y^{\prime }}{x}-\frac {y}{x^{2}}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = 0] \end {align*}
The ode can be written as \[ y^{\prime \prime } x^{2}+y^{\prime } x -y = 0 \] Which shows it is a Euler ODE.
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=\frac {1}{x}\\ q(x) &=-\frac {1}{x^{2}}\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {y^{\prime }}{x}-\frac {y}{x^{2}} = 0 \end {align*}
The domain of \(p(x)=\frac {1}{x}\) is \[
\{x <0\boldsymbol {\lor }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 \[ x^{2}(r(r-1))x^{r-2}+x r x^{r-1}-x^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}+r\,x^{r}-x^{r} = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )+r-1 = 0 \] Or \[ r^{2}-1 = 0 \tag {1} \]
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\\ 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 \[ y = \frac {c_{1}}{x}+c_{2} x \] Initial
conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1}}{x}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1}}{x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = -c_{1} +c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&={\frac {1}{2}}\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. In normal form the ode \begin {align*} y^{\prime \prime } x^{2}+y^{\prime } x -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 {1}{x}\\ q \left (x \right )&=-\frac {1}{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 {1}{x}d x \right )}d x\\ &= \int e^{-\ln \left (x \right )} \,dx\\ &= \int \frac {1}{x}d x\\ &= \ln \left (x \right )\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 {1}{x^{2}}}{\frac {1}{x^{2}}}\\ &= -1\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 )-y \left (\tau \right )&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous
ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=-1\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting
this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }-{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2)
throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}-1 = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots
determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=-1\) into the
above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-1\right )}\\ &= \pm 1 \end {align*}
Hence \begin{align*}
\lambda _1 &= + 1 \\
\lambda _2 &= - 1 \\
\end{align*} Which simplifies to \begin{align*}
\lambda _1 &= 1 \\
\lambda _2 &= -1 \\
\end{align*} Since roots are real and distinct, then the solution is \begin{align*}
y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\
y \left (\tau \right ) &= c_{1} e^{\left (1\right )\tau } +c_{2} e^{\left (-1\right )\tau } \\
\end{align*}
Or \[
y \left (\tau \right ) =c_{1} {\mathrm e}^{\tau }+c_{2} {\mathrm e}^{-\tau }
\] The above solution is now transformed back to \(y\) using (6) which results in
\begin {align*} y &= \frac {c_{1} x^{2}+c_{2}}{x} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1} x^{2}+c_{2}}{x} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{1} -\frac {c_{1} x^{2}+c_{2}}{x^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = c_{1} -c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&={\frac {1}{2}}\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. In normal form the ode \begin {align*} y^{\prime \prime } x^{2}+y^{\prime } x -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 {1}{x}\\ q \left (x \right )&=-\frac {1}{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 {-\frac {1}{x^{2}}}}{c}\tag {6} \\ \tau '' &= \frac {1}{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 {1}{c \sqrt {-\frac {1}{x^{2}}}\, x^{3}}+\frac {1}{x}\frac {\sqrt {-\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {-\frac {1}{x^{2}}}}{c}\right )^2} \\ &=0 \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 )+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 ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {-\frac {1}{x^{2}}}d x}{c}\\ &= \frac {\sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{c} \end {align*}
Substituting the above into the solution obtained gives \[
y = \frac {\left (i c_{2} +c_{1} \right ) x^{2}-i c_{2} +c_{1}}{2 x}
\] Initial conditions are used to solve
for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {\left (i c_{2} +c_{1} \right ) x^{2}-i c_{2} +c_{1}}{2 x} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = i c_{2} +c_{1} -\frac {\left (i c_{2} +c_{1} \right ) x^{2}-i c_{2} +c_{1}}{2 x^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = i c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. In normal form the ode \begin {align*} y^{\prime \prime } x^{2}+y^{\prime } x -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 {1}{x}\\ q \left (x \right )&=-\frac {1}{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 {n}{x^{2}}-\frac {1}{x^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\frac {3 v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {3 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 {3 u \left (x \right )}{x} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {3 u}{x} \end {align*}
Where \(f(x)=-\frac {3}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {3}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {3}{x} \,d x}\\ \ln \left (u \right )&=-3 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-3 \ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x^{3}} \end {align*}
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}}{2 x^{2}}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_{1}}{2 x^{2}}+c_{2} \right ) x\\ &= \left (-\frac {c_{1}}{2 x^{2}}+c_{2} \right ) x\\ \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \left (-\frac {c_{1}}{2 x^{2}}+c_{2} \right ) x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = -\frac {c_{1}}{2}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1}}{2}+c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } x^{2}+y^{\prime } x -y\right )d x &= 0 \\ x^{2} y^{\prime }-y x = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x}\\ q(x) &=\frac {c_{1}}{x^{2}} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{x} = \frac {c_{1}}{x^{2}} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {1}{x}d x} \\
&= \frac {1}{x} \\
\end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{x}\right ) &= \left (\frac {1}{x}\right ) \left (\frac {c_{1}}{x^{2}}\right )\\ \mathrm {d} \left (\frac {y}{x}\right ) &= \left (\frac {c_{1}}{x^{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{x} &= \int {\frac {c_{1}}{x^{3}}\,\mathrm {d} x}\\ \frac {y}{x} &= -\frac {c_{1}}{2 x^{2}} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{x}\) results in \begin {align*} y &= -\frac {c_{1}}{2 x}+c_{2} x \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\frac {c_{1}}{2 x}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = -\frac {c_{1}}{2}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1}}{2}+c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation
\begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which
reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode
which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution
is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= x^{2}\\ B &= x\\ C &= -1\\ F &= 0 \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{2}\right ) \left (0\right ) + \left (x\right ) \left (1\right ) + \left (-1\right ) \left (x\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} x^{3} v'' +\left ( 3 x^{2}\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} x^{2} \left (u^{\prime }\left (x \right ) x +3 u \left (x \right )\right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {3 u}{x} \end {align*}
Where \(f(x)=-\frac {3}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {3}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {3}{x} \,d x}\\ \ln \left (u \right )&=-3 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-3 \ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x^{3}} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=\frac {c_{1}}{x^{3}} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{1}}{x^{3}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (x\right ) \left (-\frac {c_{1}}{2 x^{2}}+c_{2}\right ) \\ &= \left (-\frac {c_{1}}{2 x^{2}}+c_{2} \right ) x \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \left (-\frac {c_{1}}{2 x^{2}}+c_{2} \right ) x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = -\frac {c_{1}}{2}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1}}{2}+c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. Writing the ode as \[
y^{\prime \prime } x^{2}+y^{\prime } x -y = 0
\] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } x^{2}+y^{\prime } x -y\right )d x &= 0 \\ x^{2} y^{\prime }-y x = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x}\\ q(x) &=\frac {c_{1}}{x^{2}} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{x} = \frac {c_{1}}{x^{2}} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {1}{x}d x} \\
&= \frac {1}{x} \\
\end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{x}\right ) &= \left (\frac {1}{x}\right ) \left (\frac {c_{1}}{x^{2}}\right )\\ \mathrm {d} \left (\frac {y}{x}\right ) &= \left (\frac {c_{1}}{x^{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{x} &= \int {\frac {c_{1}}{x^{3}}\,\mathrm {d} x}\\ \frac {y}{x} &= -\frac {c_{1}}{2 x^{2}} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{x}\) results in \begin {align*} y &= -\frac {c_{1}}{2 x}+c_{2} x \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\frac {c_{1}}{2 x}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = -\frac {c_{1}}{2}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1}}{2}+c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. Writing the ode as \begin {align*} y^{\prime \prime } x^{2}+y^{\prime } x -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 &= x\tag {3} \\ C &= -1 \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 {3}{4 x^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 3\\ t &= 4 x^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {3}{4 x^{2}}\right ) z(x)\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) \\ &= 2 - 0 \\ &= 2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x^{2}\).
There is a pole at \(x=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then
necessary conditions for case two are met. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin {align*} L &= [1, 2, 4, 6, 12] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[
r = \frac {3}{4 x^{2}}
\] For the pole at \(x=0\) let \(b\)
be the coefficient of \(\frac {1}{ x^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence
\begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{2}} \end {alignat*}
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from \begin {alignat*} {2} r &= \frac {s}{t} &&= \frac {3}{4 x^{2}} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {3}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{2}} \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is \[ r=\frac {3}{4 x^{2}} \]
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \)
and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative
integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until
such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = -{\frac {1}{2}}\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= -{\frac {1}{2}} - \left ( -{\frac {1}{2}} \right ) \\ &= 0 \end {align*}
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {1}{2 x} + (-) \left ( 0 \right ) \\ &= -\frac {1}{2 x}\\ &= -\frac {1}{2 x} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree
\(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}
Let \begin {align*} p(x) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (-\frac {1}{2 x}\right ) \left (0\right ) + \left ( \left (\frac {1}{2 x^{2}}\right ) + \left (-\frac {1}{2 x}\right )^2 - \left (\frac {3}{4 x^{2}}\right ) \right ) &= 0\\ 0 = 0 \end {align*}
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is
\begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int -\frac {1}{2 x}d x}\\ &= \frac {1}{\sqrt {x}} \end {align*}
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 {x}{x^{2}} \,dx} \\
&= z_1 e^{-\frac {\ln \left (x \right )}{2}} \\
&= z_1 \left (\frac {1}{\sqrt {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 {x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-\ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\frac {x^{2}}{2}\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 (\frac {x^{2}}{2}\right )\right ) \\
\end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1}}{x}+\frac {c_{2} x}{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = c_{1} +\frac {c_{2}}{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1}}{x^{2}}+\frac {c_{2}}{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = -c_{1} +\frac {c_{2}}{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&={\frac {1}{2}}\\ c_{2}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] 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) &= x\\ r(x) &= -1\\ s(x) &= 0 \end {align*}
Hence \begin {align*} p''(x) &= 2\\ q'(x) &= 1 \end {align*}
Therefore (1) becomes \begin {align*} 2- \left (1\right ) + \left (-1\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 }-y x&=c_{1} \end {align*}
We now have a first order ode to solve which is \begin {align*} x^{2} y^{\prime }-y x = c_{1} \end {align*}
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x}\\ q(x) &=\frac {c_{1}}{x^{2}} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{x} = \frac {c_{1}}{x^{2}} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {1}{x}d x} \\
&= \frac {1}{x} \\
\end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{x}\right ) &= \left (\frac {1}{x}\right ) \left (\frac {c_{1}}{x^{2}}\right )\\ \mathrm {d} \left (\frac {y}{x}\right ) &= \left (\frac {c_{1}}{x^{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{x} &= \int {\frac {c_{1}}{x^{3}}\,\mathrm {d} x}\\ \frac {y}{x} &= -\frac {c_{1}}{2 x^{2}} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{x}\) results in \begin {align*} y &= -\frac {c_{1}}{2 x}+c_{2} x \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\frac {c_{1}}{2 x}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives
\begin {align*} 1 = -\frac {c_{1}}{2}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1}}{2 x^{2}}+c_{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1}}{2}+c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}+1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}+1}{2 x} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}+1}{2 x}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x -y=0, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y^{\prime }}{x}+\frac {y}{x^{2}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y^{\prime }}{x}-\frac {y}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x -y=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} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}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}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right ) x^{2}+\frac {d}{d t}y \left (t \right )-y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y \left (t \right )-y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\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 )={\mathrm e}^{t} \\ \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} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {c_{1}}{x}+c_{2} x \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {c_{1}}{x}+c_{2} x \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {c_{1}}{x}+c_{2} x \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1}}{x^{2}}+c_{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {1}{2}, c_{2} =\frac {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {1}{2 x}+\frac {x}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {1}{2 x}+\frac {x}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 13
\[
y \left (x \right ) = \frac {1}{2 x}+\frac {x}{2}
\]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 17
\[
y(x)\to \frac {x^2+1}{2 x}
\]
12.1.2 Solving as second order euler ode ode
12.1.3 Solving as second order change of variable on x method 2 ode
12.1.4 Solving as second order change of variable on x method 1 ode
12.1.5 Solving as second order change of variable on y method 2 ode
12.1.6 Solving as second order integrable as is ode
12.1.7 Solving as second order ode non constant coeff transformation on B ode
12.1.8 Solving as type second_order_integrable_as_is (not using ABC version)
12.1.9 Solving using Kovacic algorithm
pole \(c\) location
pole order
\([\sqrt r]_c\)
\(\alpha _c^{+}\)
\(\alpha _c^{-}\)
\(0\) \(2\) \(0\) \(\frac {3}{2}\) \(-{\frac {1}{2}}\)
Order of \(r\) at \(\infty \)
\([\sqrt r]_\infty \)
\(\alpha _\infty ^{+}\)
\(\alpha _\infty ^{-}\) \(2\)
\(0\) \(\frac {3}{2}\) \(-{\frac {1}{2}}\)
12.1.10 Solving as exact linear second order ode ode
12.1.11 Maple step by step solution
`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`
dsolve([diff(y(x),x$2)+1/x*diff(y(x),x)-1/x^2*y(x)=0,y(1) = 1, D(y)(1) = 0],y(x), singsol=all)
DSolve[{y''[x]+1/x*y'[x]-1/x^2*y[x]==0,{y[1]==1,y'[1]==0}},y[x],x,IncludeSingularSolutions -> True]