problem 13.6 (f)

Internal problem ID [13511]
Internal ﬁle name [OUTPUT/12683_Friday_February_16_2024_12_10_53_AM_19995377/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending ﬁrst order concepts. Additional exercises page 259
Problem number: 13.6 (f).
ODE order: 3.
ODE degree: 1.

\[ \boxed {x y^{\prime \prime \prime }+2 y^{\prime \prime }=6 x} \] With initial conditions \begin {align*} [y \left (1\right ) = 2, y^{\prime }\left (1\right ) = 1, y^{\prime \prime }\left (1\right ) = 4] \end {align*}

Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} v^{\prime \prime }\left (x \right ) x +2 v^{\prime }\left (x \right ) = 0 \end {align*}

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (v^{\prime \prime }\left (x \right ) x +2 v^{\prime }\left (x \right )\right )d x &= 0 \\ v^{\prime }\left (x \right ) x +v \left (x \right ) = c_{1} \end {align*}

Which is now solved for \(v \left (x \right )\). In canonical form the ODE is \begin {align*} v' &= F(x,v)\\ &= f( x) g(v)\\ &= \frac {-v +c_{1}}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(v)=-v +c_{1}\). Integrating both sides gives \begin{align*} \frac {1}{-v +c_{1}} \,dv &= \frac {1}{x} \,d x \\ \int { \frac {1}{-v +c_{1}} \,dv} &= \int {\frac {1}{x} \,d x} \\ -\ln \left (v -c_{1} \right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{v -c_{1}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} \frac {1}{v -c_{1}} &= c_{3} x \end {align*}

Which simpliﬁes to \[ v \left (x \right ) = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \] But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x}\). Integrating both sides gives \begin {align*} y &= \int { \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x}\,\mathop {\mathrm {d}x}}\\ &= c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4} \end {align*}

This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ x y^{\prime \prime \prime }+2 y^{\prime \prime } = 0 \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coeﬃcient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the ﬁrst step is to ﬁnd the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3\\ y_1'&y_2'&y_3'\\ y_1''&y_2''&y_3''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{ccc} 1 & x & \ln \left (x \right ) \\ 0 & 1 & \frac {1}{x} \\ 0 & 0 & -\frac {1}{x^{2}} \end {array}\right ] \\ |W| &= -\frac {1}{x^{2}} \end {align*}

The determinant simpliﬁes to \begin {align*} |W| &= -\frac {1}{x^{2}} \end {align*}

Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{cc} x & \ln \left (x \right ) \\ 1 & \frac {1}{x} \end {array}\right ] \\ &= 1-\ln \left (x \right ) \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} 1 & \ln \left (x \right ) \\ 0 & \frac {1}{x} \end {array}\right ] \\ &= \frac {1}{x} \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} 1 & x \\ 0 & 1 \end {array}\right ] \\ &= 1 \end {align*}

Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{3-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (6 x\right ) \left (1-\ln \left (x \right )\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= \int { \frac {6 x \left (1-\ln \left (x \right )\right )}{-\frac {1}{x}} \, dx}\\ &= \int {\left (-6 x^{2} \left (1-\ln \left (x \right )\right )\right ) \, dx}\\ &= -\frac {8 x^{3}}{3}+2 \ln \left (x \right ) x^{3} \end {align*}

\begin {align*} U_2 &= (-1)^{3-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (6 x\right ) \left (\frac {1}{x}\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= - \int { \frac {6}{-\frac {1}{x}} \, dx}\\ &= - \int {\left (-6 x\right ) \, dx}\\ &= 3 x^{2} \end {align*}

\begin {align*} U_3 &= (-1)^{3-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (6 x\right ) \left (1\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= \int { \frac {6 x}{-\frac {1}{x}} \, dx}\\ &= \int {\left (-6 x^{2}\right ) \, dx}\\ &= -2 x^{3} \end {align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Hence \begin {equation*} \begin {split} y_p &= \left (-\frac {8 x^{3}}{3}+2 \ln \left (x \right ) x^{3}\right ) \\ &+\left (3 x^{2}\right ) \left (x\right ) \\ &+\left (-2 x^{3}\right ) \left (\ln \left (x \right )\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = \frac {x^{3}}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4}\right ) + \left (\frac {x^{3}}{3}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4} +\frac {x^{3}}{3} \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 \(y = 2\) and \(x = 1\) in the above gives \begin {align*} 2 = c_{1} +c_{4} +\frac {1}{3}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} +\frac {{\mathrm e}^{-c_{2}}}{c_{3} x}+x^{2} \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = \frac {c_{3} c_{1} +{\mathrm e}^{-c_{2}}+c_{3}}{c_{3}}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = -\frac {{\mathrm e}^{-c_{2}}}{c_{3} x^{2}}+2 x \end {align*}

substituting \(y^{\prime \prime } = 4\) and \(x = 1\) in the above gives \begin {align*} 4 = \frac {-{\mathrm e}^{-c_{2}}+2 c_{3}}{c_{3}}\tag {3A} \end {align*}

Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=2\\ c_{2}&=-\ln \left (2\right )+\ln \left (-\frac {1}{c_{3}}\right )\\ c_{4}&=-{\frac {1}{3}} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \frac {x^{3}}{3}-2 \ln \left (x \right )-\frac {1}{3}+2 x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{3}}{3}-2 \ln \left (x \right )-\frac {1}{3}+2 x \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {x^{3}}{3}-2 \ln \left (x \right )-\frac {1}{3}+2 x \] Veriﬁed OK.

8.40.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x y^{\prime \prime \prime }+2 y^{\prime \prime }=6 x , y \left (1\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
<- high order exact linear fully integrable successful`

\[ y \left (x \right ) = \frac {x^{3}}{3}-2 \ln \left (x \right )+2 x -\frac {1}{3} \]

8.40 problem 13.6 (f)

8.40.1 Maple step by step solution