7.8 problem 18

Internal problem ID [6695]
Internal file name [OUTPUT/5943_Sunday_June_05_2022_04_03_12_PM_67799813/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.4.1 DERIVATIVES OF A TRANSFORM. Page 309
Problem number: 18.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "kovacic"

Maple gives the following as the ode type

[[_2nd_order, _with_linear_symmetries]]

Unable to solve or complete the solution.

\[ \boxed {2 y^{\prime \prime }+t y^{\prime }-2 y=10} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

7.8.1 Existence and uniqueness analysis

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 {t}{2}\\ q(t) &=-1\\ F &=5 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {t y^{\prime }}{2}-y = 5 \end {align*}

The domain of \(p(t)=\frac {t}{2}\) is \[ \{-\infty

7.8.2 Solving using Kovacic algorithm

Writing the ode as \begin {align*} 2 y^{\prime \prime }+t y^{\prime }-2 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 &= 2 \\ B &= t\tag {3} \\ C &= -2 \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 {t^{2}+20}{16}\tag {6} \end {align*}

Comparing the above to (5) shows that \begin {align*} s &= t^{2}+20\\ t &= 16 \end {align*}

Therefore eq. (4) becomes \begin {align*} z''(t) &= \left ( \frac {t^{2}}{16}+\frac {5}{4}\right ) z(t)\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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 50: 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 - 2 \\ &= -2 \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 \(-2\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}

Attempting to find a solution using case \(n=1\).

Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -2\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {2}{2} &&= 1 \end {alignat*}

\([\sqrt r]_\infty \) is the sum of terms involving \(t^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i t^i \\ &= \sum _{i=0}^{1} a_i t^i \tag {8} \end {align*}

Let \(a\) be the coefficient of \(t^v=t^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {t}{4}+\frac {5}{2 t}-\frac {25}{2 t^{3}}+\frac {125}{t^{5}}-\frac {3125}{2 t^{7}}+\frac {21875}{t^{9}}-\frac {328125}{t^{11}}+\frac {5156250}{t^{13}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = {\frac {1}{4}} \] From Eq. (9) the sum up to \(v=1\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{1} a_i t^i \\ &= \frac {t}{4} \tag {10} \end {align*}

Now we need to find \(b\), where \(b\) be the coefficient of \(t^{v-1} = t^{0}=1\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = \frac {t^{2}}{16} \] This shows that the coefficient of \(1\) in the above is \(0\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=1\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {t^{2}+20}{16} \\ &= Q + \frac {R}{16} \\ &= \left (\frac {t^{2}}{16}+\frac {5}{4}\right ) + \left ( 0\right ) \\ &= \frac {t^{2}}{16}+\frac {5}{4} \end {align*}

We see that the coefficient of the term \(\frac {1}{t}\) in the quotient is \(\frac {5}{4}\). Now \(b\) can be found. \begin {align*} b &= \left ({\frac {5}{4}}\right )-\left (0\right )\\ &= {\frac {5}{4}} \end {align*}

Hence \begin {alignat*} {3} [\sqrt r]_\infty &= \frac {t}{4}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {{\frac {5}{4}}}{{\frac {1}{4}}} - 1 \right ) &&= 2\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {{\frac {5}{4}}}{{\frac {1}{4}}} - 1 \right ) &&= -3 \end {alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {t^{2}}{16}+\frac {5}{4} \]

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(-2\)	\(\frac {t}{4}\)	\(2\)	\(-3\)

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 \(\alpha _\infty ^{+} = 2\), and since there are no poles, then \begin {align*} d &= \alpha _\infty ^{+} \\ &= 2 \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)}}{t-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}

Substituting the above values in the above results in \begin {align*} \omega &= (+) [\sqrt r]_\infty \\ &= 0 + \left ( \frac {t}{4} \right ) \\ &= \frac {t}{4}\\ &= \frac {t}{4} \end {align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(t)\) of degree \(d=2\) to solve the ode. The polynomial \(p(t)\) 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(t) &= t^{2}+a_{1} t +a_{0}\tag {2A} \end {align*}

Substituting the above in eq. (1A) gives \begin {align*} \left (2\right ) + 2 \left (\frac {t}{4}\right ) \left (2 t +a_{1}\right ) + \left ( \left ({\frac {1}{4}}\right ) + \left (\frac {t}{4}\right )^2 - \left (\frac {t^{2}}{16}+\frac {5}{4}\right ) \right ) &= 0\\ 2-\frac {a_{1} t}{2}-a_{0} = 0 \end {align*}

Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives \[ \{a_{0} = 2, a_{1} = 0\} \] Substituting these coefficients in \(p(t)\) in eq. (2A) results in \begin {align*} p(t) &= t^{2}+2 \end {align*}

Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(t) &= p e^{ \int \omega \,dt}\\ & = \left (t^{2}+2\right ) {\mathrm e}^{\int \frac {t}{4}d t}\\ & = \left (t^{2}+2\right ) {\mathrm e}^{\frac {t^{2}}{8}}\\ & = \left (t^{2}+2\right ) {\mathrm e}^{\frac {t^{2}}{8}} \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} \,dt} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {t}{2} \,dt} \\ &= z_1 e^{-\frac {t^{2}}{8}} \\ &= z_1 \left ({\mathrm e}^{-\frac {t^{2}}{8}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = t^{2}+2 \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {t}{2} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{-\frac {t^{2}}{4}}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t\right ) \\ \end{align*} Therefore the solution is

\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (t^{2}+2\right ) + c_{2} \left (t^{2}+2\left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t\right )\right ) \\ \end{align*} This is second order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(t) + B y'(t) + C y(t) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). \(y_h\) is the solution to \[ 2 y^{\prime \prime }+t y^{\prime }-2 y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = c_{1} \left (t^{2}+2\right )+c_{2} \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(t\) as well. Let \begin{equation} \tag{1} y_p(t) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= t^{2}+2 \\ y_2 &= \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(t)}{a W(t)} \\ \tag{3} u_2 &= \int \frac {y_1 f(t)}{a W(t)} \\ \end{align*} Where \(W(t)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} t^{2}+2 & \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) \\ \frac {d}{dt}\left (t^{2}+2\right ) & \frac {d}{dt}\left (\left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} t^{2}+2 & \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) \\ 2 t & 2 t \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+\frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{t^{2}+2} \end {vmatrix} \] Therefore \[ W = \left (t^{2}+2\right )\left (2 t \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+\frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{t^{2}+2}\right ) - \left (\left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )\right )\left (2 t\right ) \] Which simplifies to \[ W = {\mathrm e}^{-\frac {t^{2}}{4}} \] Which simplifies to \[ W = {\mathrm e}^{-\frac {t^{2}}{4}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {10 \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )}{2 \,{\mathrm e}^{-\frac {t^{2}}{4}}}\,dt \] Which simplifies to \[ u_1 = - \int 5 \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) {\mathrm e}^{\frac {t^{2}}{4}}d t \] Hence \[ u_1 = -\left (\int _{0}^{t}5 \left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {10 t^{2}+20}{2 \,{\mathrm e}^{-\frac {t^{2}}{4}}}\,dt \] Which simplifies to \[ u_2 = \int 5 \left (t^{2}+2\right ) {\mathrm e}^{\frac {t^{2}}{4}}d t \] Hence \[ u_2 = 10 t \,{\mathrm e}^{\frac {t^{2}}{4}} \] Which simplifies to \begin{align*} u_1 &= -5 \left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right ) \\ u_2 &= 10 t \,{\mathrm e}^{\frac {t^{2}}{4}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(t) = -5 \left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right ) \left (t^{2}+2\right )+10 t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) \] Which simplifies to \[ y_p(t) = 10 \left (t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}\right ) \left (t^{2}+2\right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} \left (t^{2}+2\right )+c_{2} \left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )\right ) + \left (10 \left (t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}\right ) \left (t^{2}+2\right )\right ) \\ \end{align*} Which simplifies to \[ y = \left (t^{2}+2\right ) \left (c_{2} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+c_{1} \right )+10 \left (t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}\right ) \left (t^{2}+2\right ) \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \left (t^{2}+2\right ) \left (c_{2} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+c_{1} \right )+10 \left (t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}\right ) \left (t^{2}+2\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 \(t = 0\) in the above gives \begin {align*} 0 = \munderset {t \rightarrow 0}{\operatorname {lim}}10 \left (t^{2}+2\right ) \left (-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}+\left (t \,{\mathrm e}^{\frac {t^{2}}{4}}+\frac {c_{2}}{10}\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+\frac {c_{1}}{10}\right )\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 t \left (c_{2} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+c_{1} \right )+\frac {c_{2} {\mathrm e}^{-\frac {t^{2}}{4}}}{t^{2}+2}+10 \left ({\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+\frac {t^{2} {\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )}{2}+\frac {t \,{\mathrm e}^{\frac {t^{2}}{4}} {\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}-\frac {\left (t^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right ) {\mathrm e}^{\frac {t^{2}}{4}}}{2}\right ) \left (t^{2}+2\right )+20 \left (t \,{\mathrm e}^{\frac {t^{2}}{4}} \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )-\frac {\left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )}{2}\right ) t \end {align*}

substituting \(y^{\prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = \munderset {t \rightarrow 0}{\operatorname {lim}}\frac {\left (-10 t^{3}-20 t \right ) \left (\int _{0}^{t}\left (\alpha ^{2}+2\right ) \left (\int \frac {{\mathrm e}^{-\frac {\alpha ^{2}}{4}}}{\left (\alpha ^{2}+2\right )^{2}}d \alpha \right ) {\mathrm e}^{\frac {\alpha ^{2}}{4}}d \alpha \right )+20 \left (t^{2}+2\right ) \left (t \,{\mathrm e}^{\frac {t^{2}}{4}}+\frac {c_{2}}{10}\right ) t \left (\int \frac {{\mathrm e}^{-\frac {t^{2}}{4}}}{\left (t^{2}+2\right )^{2}}d t \right )+c_{2} {\mathrm e}^{-\frac {t^{2}}{4}}+2 t \left (t^{2} c_{1} +2 c_{1} +5\right )}{t^{2}+2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.

Verification of solutions N/A

Maple trace Kovacic algorithm successful

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   checking if the LODE is of Euler type 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying a Liouvillian solution using Kovacics algorithm 
      A Liouvillian solution exists 
      Reducible group (found an exponential solution) 
      Group is reducible, not completely reducible 
      Solution has integrals. Trying a special function solution free of integrals... 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         <- Kummer successful 
      <- special function solution successful 
         -> Trying to convert hypergeometric functions to elementary form... 
         <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals 
      <- Kovacics algorithm successful 
<- solving first the homogeneous part of the ODE successful`
 

✓ Solution by Maple

Time used: 0.062 (sec). Leaf size: 9

dsolve([2*diff(y(t),t$2)+t*diff(y(t),t)-2*y(t)=10,y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {5 t^{2}}{2} \]

✓ Solution by Mathematica

Time used: 0.02 (sec). Leaf size: 10

DSolve[{y''[t]+t*y'[t]-2*y[t]==10,{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 5 t^2 \]