Internal problem ID [14089]
Internal file name [OUTPUT/13770_Saturday_March_02_2024_02_49_23_PM_5762845/index.tex
]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 1. Introduction to Differential Equations. Exercises 1.1, page 10
Problem number: 53.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_euler_ode", "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_1", "second_order_change_of_variable_on_y_method_2", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = -1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}
Where here \begin {align*} p(t) &=-\frac {12}{t}\\ q(t) &=\frac {42}{t^{2}}\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-\frac {12 y^{\prime }}{t}+\frac {42 y}{t^{2}} = 0 \end {align*}
The domain of \(p(t)=-\frac {12}{t}\) is \[
\{t <0\boldsymbol {\lor }0
This is Euler second order ODE. Let the solution be \(y = t^r\), then \(y'=r t^{r-1}\) and \(y''=r(r-1) t^{r-2}\). Substituting these back
into the given ODE gives \[ t^{2}(r(r-1))t^{r-2}-12 t r t^{r-1}+42 t^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )t^{r}-12 r\,t^{r}+42 t^{r} = 0 \] Since \(t^{r}\neq 0\) then dividing throughout by \(t^{r}\) gives \[ r \left (r -1\right )-12 r+42 = 0 \] Or \[ r^{2}-13 r +42 = 0 \tag {1} \]
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 &= 6\\ r_2 &= 7 \end {align*}
Since the roots are real and distinct, then the general solution is \[ y= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = t^{r_1}\) and \(y_2 = t^{r_2} \). Hence \[ y = c_{2} t^{7}+c_{1} t^{6} \] Initial
conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} t^{7}+c_{1} t^{6} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 7 c_{2} t^{6}+6 c_{1} t^{5} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = 7 c_{2} +6 c_{1}\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}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -t^{7}+t^{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -t^{7}+t^{6} \\
\end{align*} Verification of solutions
\[
y = -t^{7}+t^{6}
\] Verified OK. The ode satisfies this form \[ y^{\prime \prime }+p \left (t \right ) y^{\prime }+\frac {\left (p \left (t \right )^{2}+p^{\prime }\left (t \right )\right ) y}{2} = f \left (t \right ) \] Where \( p(t) = -\frac {12}{t}\). Therefore, there is an integrating factor given by
\begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int -\frac {12}{t} \, dx} \\ &= \frac {1}{t^{6}} \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}{t^{6}} \right )'' &= 0 \\
\end{align*} Integrating once gives \[ \left ( \frac {y}{t^{6}} \right )' = c_{1} \] Integrating again gives \[ \left ( \frac {y}{t^{6}} \right ) = c_{1} t +c_{2} \]
Hence the solution is \begin{align*}
y &= \frac {c_{1} t +c_{2}}{\frac {1}{t^{6}}} \\
\end{align*} Or \[
y = c_{1} t^{7}+c_{2} t^{6}
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = c_{1} t^{7}+c_{2} t^{6} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 7 c_{1} t^{6}+6 c_{2} t^{5} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = 7 c_{1} +6 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}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -t^{7}+t^{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -t^{7}+t^{6} \\
\end{align*} Verification of solutions
\[
y = -t^{7}+t^{6}
\] Verified OK. In normal form the ode \begin {align*} t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (t \right )&=-\frac {12}{t}\\ q \left (t \right )&=\frac {42}{t^{2}} \end {align*}
Applying change of variables \(\tau = g \left (t \right )\) 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 (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (t \right )d t \right )}d t\\ &= \int {\mathrm e}^{-\left (\int -\frac {12}{t}d t \right )}d t\\ &= \int e^{12 \ln \left (t \right )} \,dt\\ &= \int t^{12}d t\\ &= \frac {t^{13}}{13}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &= \frac {\frac {42}{t^{2}}}{t^{24}}\\ &= \frac {42}{t^{26}}\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 {42 y \left (\tau \right )}{t^{26}}&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} \frac {42}{t^{26}}&=\frac {42}{169 \tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {42 y \left (\tau \right )}{169 \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 169 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+42 y \left (\tau \right ) = 0 \] Which shows it is a
Euler ODE. 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 \[ 169 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+42 \tau ^{r} = 0 \] Simplifying gives \[ 169 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+42 \tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing
throughout by \(\tau ^{r}\) gives \[ 169 r \left (r -1\right )+0+42 = 0 \] Or \[ 169 r^{2}-169 r +42 = 0 \tag {1} \] 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 {6}{13}}\\ r_2 &= {\frac {7}{13}} \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 \[ y \left (\tau \right ) = c_{1} \tau ^{\frac {6}{13}}+c_{2} \tau ^{\frac {7}{13}} \] The above solution is now transformed back to \(y\) using (6) which results in
\begin {align*} y &= \frac {c_{1} 13^{\frac {7}{13}} \left (t^{13}\right )^{\frac {6}{13}}}{13}+\frac {c_{2} 13^{\frac {6}{13}} \left (t^{13}\right )^{\frac {7}{13}}}{13} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1} 13^{\frac {7}{13}} \left (t^{13}\right )^{\frac {6}{13}}}{13}+\frac {c_{2} 13^{\frac {6}{13}} \left (t^{13}\right )^{\frac {7}{13}}}{13} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = \frac {13^{\frac {6}{13}} \left (c_{1} 13^{\frac {1}{13}}+c_{2} \right )}{13}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {6 c_{1} 13^{\frac {7}{13}} t^{12}}{13 \left (t^{13}\right )^{\frac {7}{13}}}+\frac {7 c_{2} 13^{\frac {6}{13}} t^{12}}{13 \left (t^{13}\right )^{\frac {6}{13}}} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = \frac {\left (6 c_{1} 13^{\frac {1}{13}}+7 c_{2} \right ) 13^{\frac {6}{13}}}{13}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=13^{\frac {6}{13}}\\ c_{2}&=-13^{\frac {7}{13}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\left (t^{13}\right )^{\frac {7}{13}}+\left (t^{13}\right )^{\frac {6}{13}} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\left (t^{13}\right )^{\frac {7}{13}}+\left (t^{13}\right )^{\frac {6}{13}} \\
\end{align*} Verification of solutions
\[
y = -\left (t^{13}\right )^{\frac {7}{13}}+\left (t^{13}\right )^{\frac {6}{13}}
\] Verified OK. In normal form the ode \begin {align*} t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (t \right )&=-\frac {12}{t}\\ q \left (t \right )&=\frac {42}{t^{2}} \end {align*}
Applying change of variables \(\tau = g \left (t \right )\) 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 (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \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 {42}\, \sqrt {\frac {1}{t^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {42}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &=\frac {-\frac {\sqrt {42}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}}-\frac {12}{t}\frac {\sqrt {42}\, \sqrt {\frac {1}{t^{2}}}}{c}}{\left (\frac {\sqrt {42}\, \sqrt {\frac {1}{t^{2}}}}{c}\right )^2} \\ &=-\frac {13 c \sqrt {42}}{42} \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 {13 c \sqrt {42}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{42}+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 {13 \sqrt {42}\, c \tau }{84}} \left (c_{1} \cosh \left (\frac {\sqrt {42}\, c \tau }{84}\right )+i c_{2} \sinh \left (\frac {\sqrt {42}\, c \tau }{84}\right )\right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dt \\ &= \frac {\int \sqrt {42}\, \sqrt {\frac {1}{t^{2}}}d t}{c}\\ &= \frac {\sqrt {42}\, \sqrt {\frac {1}{t^{2}}}\, t \ln \left (t \right )}{c} \end {align*}
Substituting the above into the solution obtained gives \[
y = t^{\frac {13}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (t \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (t \right )}{2}\right )\right )
\] Initial conditions are used to solve
for the constants of integration.
Looking at the above solution \begin {align*} y = t^{\frac {13}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (t \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (t \right )}{2}\right )\right ) \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {13 t^{\frac {11}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (t \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (t \right )}{2}\right )\right )}{2}+t^{\frac {13}{2}} \left (\frac {c_{1} \sinh \left (\frac {\ln \left (t \right )}{2}\right )}{2 t}+\frac {i c_{2} \cosh \left (\frac {\ln \left (t \right )}{2}\right )}{2 t}\right ) \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = \frac {13 c_{1}}{2}+\frac {i 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}&=0\\ c_{2}&=2 i \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -2 t^{\frac {13}{2}} \sinh \left (\frac {\ln \left (t \right )}{2}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -2 t^{\frac {13}{2}} \sinh \left (\frac {\ln \left (t \right )}{2}\right ) \\
\end{align*} Verification of solutions
\[
y = -2 t^{\frac {13}{2}} \sinh \left (\frac {\ln \left (t \right )}{2}\right )
\] Verified OK. In normal form the given ode is written as \begin {align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (t \right )&=-\frac {12}{t}\\ q \left (t \right )&=\frac {42}{t^{2}} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {42}{t^{2}} - \frac {\left (-\frac {12}{t}\right )'}{2}- \frac {\left (-\frac {12}{t}\right )^2}{4} \\ &= \frac {42}{t^{2}} - \frac {\left (\frac {12}{t^{2}}\right )}{2}- \frac {\left (\frac {144}{t^{2}}\right )}{4} \\ &= \frac {42}{t^{2}} - \left (\frac {6}{t^{2}}\right )-\frac {36}{t^{2}}\\ &= 0 \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(t\) then the
transformation \begin {align*} y = v \left (t \right ) z \left (t \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (t \right )\) is given
by \begin {align*} z \left (t \right )&={\mathrm e}^{-\left (\int \frac {p \left (t \right )}{2}d t \right )}\\ &= e^{-\int \frac {-\frac {12}{t}}{2} }\\ &= t^{6}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (t \right ) t^{6}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} t^{8} v^{\prime \prime }\left (t \right ) = 0 \end {align*}
Which is now solved for \(v \left (t \right )\) Integrating twice gives the solution \[ v \left (t \right )= c_{1} t + c_{2} \] Now that \(v \left (t \right )\) is known, then
\begin {align*} y&= v \left (t \right ) z \left (t \right )\\ &= \left (c_{1} t +c_{2}\right ) \left (z \left (t \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (t \right )&= t^{6} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} t +c_{2} \right ) t^{6} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \left (c_{1} t +c_{2} \right ) t^{6} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} t^{6}+6 \left (c_{1} t +c_{2} \right ) t^{5} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = 7 c_{1} +6 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}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -t^{6} \left (t -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -t^{6} \left (t -1\right ) \\
\end{align*} Verification of solutions
\[
y = -t^{6} \left (t -1\right )
\] Verified OK. In normal form the ode \begin {align*} t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (t \right )&=-\frac {12}{t}\\ q \left (t \right )&=\frac {42}{t^{2}} \end {align*}
Applying change of variables on the depndent variable \(y = v \left (t \right ) t^{n}\) to (2) gives the following ode where
the dependent variables is \(v \left (t \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (t \right )+\left (\frac {2 n}{t}+p \right ) v^{\prime }\left (t \right )+\left (\frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q \right ) v \left (t \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (t \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (t \right )\) and \(q \left (t \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{t^{2}}-\frac {12 n}{t^{2}}+\frac {42}{t^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=7 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (t \right )+\frac {2 v^{\prime }\left (t \right )}{t}&=0 \\ v^{\prime \prime }\left (t \right )+\frac {2 v^{\prime }\left (t \right )}{t}&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (t \right ) = v^{\prime }\left (t \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (t \right )+\frac {2 u \left (t \right )}{t} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (t \right )\). In canonical form the ODE is \begin {align*} u' &= F(t,u)\\ &= f( t) g(u)\\ &= -\frac {2 u}{t} \end {align*}
Where \(f(t)=-\frac {2}{t}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2}{t} \,d t\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2}{t} \,d t}\\ \ln \left (u \right )&=-2 \ln \left (t \right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (t \right )+c_{1}}\\ &=\frac {c_{1}}{t^{2}} \end {align*}
Now that \(u \left (t \right )\) is known, then \begin {align*} v^{\prime }\left (t \right )&= u \left (t \right )\\ v \left (t \right )&= \int u \left (t \right )d t +c_{2}\\ &= -\frac {c_{1}}{t}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (t \right ) t^{n}\\ &= \left (-\frac {c_{1}}{t}+c_{2} \right ) t^{7}\\ &= \left (c_{2} t -c_{1} \right ) t^{6}\\ \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}}{t}+c_{2} \right ) t^{7} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = -c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} t^{5}+7 \left (-\frac {c_{1}}{t}+c_{2} \right ) t^{6} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = -6 c_{1} +7 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}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -t^{6} \left (t -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -t^{6} \left (t -1\right ) \\
\end{align*} Verification of solutions
\[
y = -t^{6} \left (t -1\right )
\] Verified OK. Writing the ode as \begin {align*} t^{2} y^{\prime \prime }-12 t y^{\prime }+42 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 &= t^{2} \\ B &= -12 t\tag {3} \\ C &= 42 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end {align*}
Then (2) becomes \begin {align*} z''(t) = r z(t)\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''(t) &= 0 \tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation
\begin {align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \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 \(t\), then there is no need run Kovacic algorithm to obtain a solution
for transformed ode \(z''=r z\) as one solution is \[ z_1(t) = 1 \] 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} \,dt} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {-12 t}{t^{2}} \,dt} \\
&= z_1 e^{6 \ln \left (t \right )} \\
&= z_1 \left (t^{6}\right ) \\
\end{align*} Which simplifies to \[
y_1 = t^{6}
\]
The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \] Substituting gives \begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {-12 t}{t^{2}} \,dt}}{\left (y_1\right )^2} \,dt \\
&= y_1 \int \frac { e^{12 \ln \left (t \right )}}{\left (y_1\right )^2} \,dt \\
&= y_1 \left (t\right ) \\
\end{align*}
Therefore the solution is
\begin{align*}
y &= c_{1} y_1 + c_{2} y_2 \\
&= c_{1} \left (t^{6}\right ) + c_{2} \left (t^{6}\left (t\right )\right ) \\
\end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} t^{7}+c_{1} t^{6} \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 \(y = 0\) and \(t = 1\) in the above gives
\begin {align*} 0 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 7 c_{2} t^{6}+6 c_{1} t^{5} \end {align*}
substituting \(y^{\prime } = -1\) and \(t = 1\) in the above gives \begin {align*} -1 = 7 c_{2} +6 c_{1}\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}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -t^{7}+t^{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -t^{7}+t^{6} \\
\end{align*} Verification of solutions
\[
y = -t^{7}+t^{6}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y=0, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=-1\right ] \\ \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 {12 y^{\prime }}{t}-\frac {42 y}{t^{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} \\ {} & {} & y^{\prime \prime }-\frac {12 y^{\prime }}{t}+\frac {42 y}{t^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }-12 t y^{\prime }+42 y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & s =\ln \left (t \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d s}y \left (s \right )}{t} \\ {} & \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2}+s^{\prime \prime }\left (t \right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}}\right )-12 \frac {d}{d s}y \left (s \right )+42 y \left (s \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d s^{2}}y \left (s \right )-13 \frac {d}{d s}y \left (s \right )+42 y \left (s \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-13 r +42=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -6\right ) \left (r -7\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (6, 7\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (s \right )={\mathrm e}^{6 s} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (s \right )={\mathrm e}^{7 s} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (s \right )=c_{1} y_{1}\left (s \right )+c_{2} y_{2}\left (s \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (s \right )=c_{1} {\mathrm e}^{6 s}+c_{2} {\mathrm e}^{7 s} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y=c_{2} t^{7}+c_{1} t^{6} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=t^{6} \left (c_{2} t +c_{1} \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=t^{6} \left (c_{2} t +c_{1} \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=6 t^{5} \left (c_{2} t +c_{1} \right )+c_{2} t^{6} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=-1 \\ {} & {} & -1=7 c_{2} +6 c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =-1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-t^{6} \left (t -1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-t^{6} \left (t -1\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 13
\[
y \left (t \right ) = -t^{7}+t^{6}
\]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 13
\[
y(t)\to -\left ((t-1) t^6\right )
\]
1.46.2 Solving as second order euler ode ode
1.46.3 Solving as linear second order ode solved by an integrating factor ode
1.46.4 Solving as second order change of variable on x method 2 ode
1.46.5 Solving as second order change of variable on x method 1 ode
1.46.6 Solving as second order change of variable on y method 1 ode
1.46.7 Solving as second order change of variable on y method 2 ode
1.46.8 Solving using Kovacic algorithm
1.46.9 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([t^2*diff(y(t),t$2)-12*t*diff(y(t),t)+42*y(t)=0,y(1) = 0, D(y)(1) = -1],y(t), singsol=all)
DSolve[{t^2*y''[t]-12*t*y'[t]+42*y[t]==0,{y[1]==0,y'[1]==-1}},y[t],t,IncludeSingularSolutions -> True]