36.1 problem Ex 1

36.1.1 Maple step by step solution

Internal problem ID [11322]
Internal file name [OUTPUT/10308_Tuesday_December_27_2022_04_06_11_AM_15567907/index.tex]

Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers. 1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than first. Article 60. Exact equation. Integrating factor. Page 139
Problem number: Ex 1.
ODE order: 3.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_3rd_order, _missing_y]]

\[ \boxed {\left (x +2\right )^{2} y^{\prime \prime \prime }+\left (x +2\right ) y^{\prime \prime }+y^{\prime }=1} \] 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 \left (x \right )+\left (x +2\right ) v^{\prime }\left (x \right )+\left (x^{2}+4 x +4\right ) v^{\prime \prime }\left (x \right ) = 0 \end {align*}

In normal form the ode \begin {align*} v^{\prime \prime }\left (x \right ) \left (x +2\right )^{2}+\left (x +2\right ) v^{\prime }\left (x \right )+v \left (x \right )&=0 \tag {1} \end {align*}

Becomes \begin {align*} v^{\prime \prime }\left (x \right )+p \left (x \right ) v^{\prime }\left (x \right )+q \left (x \right ) v \left (x \right )&=0 \tag {2} \end {align*}

Where \begin {align*} p \left (x \right )&=\frac {1}{x +2}\\ q \left (x \right )&=\frac {1}{\left (x +2\right )^{2}} \end {align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}v \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }v \left (\tau \right )\right )+q_{1} v \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 +2}d x \right )}d x\\ &= \int e^{-\ln \left (x +2\right )} \,dx\\ &= \int \frac {1}{x +2}d x\\ &= \ln \left (x +2\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}{\left (x +2\right )^{2}}}{\frac {1}{\left (x +2\right )^{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}}v \left (\tau \right )+q_{1} v \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}v \left (\tau \right )+v \left (\tau \right )&=0 \end {align*}

The above ode is now solved for \(v \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A v''(\tau ) + B v'(\tau ) + C v(\tau ) = 0 \] Where in the above \(A=1, B=0, C=1\). Let the solution be \(v \left (\tau \right )=e^{\lambda \tau }\). 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 i \end {align*}

Hence \begin {align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end {align*}

Which simplifies to \begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*} Since roots are complex conjugate of each others, then let the roots be \[ \lambda _{1,2} = \alpha \pm i \beta \] Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as \[ v \left (\tau \right ) = e^{\alpha \tau } \left ( c_{1} \cos (\beta \tau ) + c_{2} \sin (\beta \tau ) \right ) \] Which becomes \[ v \left (\tau \right ) = e^{0}\left (c_{1} \cos \left (\tau \right )+c_{2} \sin \left (\tau \right )\right ) \] Or \[ v \left (\tau \right ) = c_{1} \cos \left (\tau \right )+c_{2} \sin \left (\tau \right ) \] The above solution is now transformed back to \(v \left (x \right )\) using (6) which results in \begin {align*} v \left (x \right ) &= c_{1} \cos \left (\ln \left (x +2\right )\right )+c_{2} \sin \left (\ln \left (x +2\right )\right ) \end {align*}

But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = c_{1} \cos \left (\ln \left (x +2\right )\right )+c_{2} \sin \left (\ln \left (x +2\right )\right )\). Integrating both sides gives \begin {align*} y &= \int { c_{1} \cos \left (\ln \left (x +2\right )\right )+c_{2} \sin \left (\ln \left (x +2\right )\right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (-c_{2} +c_{1} \right ) \left (x +2\right ) \cos \left (\ln \left (x +2\right )\right )}{2}+\frac {\left (c_{1} +c_{2} \right ) \left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2}+2 c_{1} -2 c_{2} +c_{3} \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 \[ y^{\prime }+\left (x +2\right ) y^{\prime \prime }+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime } = 0 \] 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 & \frac {\left (-2-x \right ) \cos \left (\ln \left (x +2\right )\right )}{2}-2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} & \frac {\cos \left (\ln \left (x +2\right )\right ) \left (x +2\right )}{2}+2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} \\ 0 & \sin \left (\ln \left (x +2\right )\right ) & \cos \left (\ln \left (x +2\right )\right ) \\ 0 & \frac {\cos \left (\ln \left (x +2\right )\right )}{x +2} & -\frac {\sin \left (\ln \left (x +2\right )\right )}{x +2} \end {array}\right ] \\ |W| &= -\frac {\cos \left (\ln \left (x +2\right )\right )^{2}+\sin \left (\ln \left (x +2\right )\right )^{2}}{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} \frac {\left (-2-x \right ) \cos \left (\ln \left (x +2\right )\right )}{2}-2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} & \frac {\cos \left (\ln \left (x +2\right )\right ) \left (x +2\right )}{2}+2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} \\ \sin \left (\ln \left (x +2\right )\right ) & \cos \left (\ln \left (x +2\right )\right ) \end {array}\right ] \\ &= -\frac {x}{2}-1-2 \cos \left (\ln \left (x +2\right )\right )-2 \sin \left (\ln \left (x +2\right )\right ) \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} 1 & \frac {\cos \left (\ln \left (x +2\right )\right ) \left (x +2\right )}{2}+2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} \\ 0 & \cos \left (\ln \left (x +2\right )\right ) \end {array}\right ] \\ &= \cos \left (\ln \left (x +2\right )\right ) \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} 1 & \frac {\left (-2-x \right ) \cos \left (\ln \left (x +2\right )\right )}{2}-2+\frac {\left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2} \\ 0 & \sin \left (\ln \left (x +2\right )\right ) \end {array}\right ] \\ &= \sin \left (\ln \left (x +2\right )\right ) \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 (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \left (-\frac {x}{2}-1-2 \cos \left (\ln \left (x +2\right )\right )-2 \sin \left (\ln \left (x +2\right )\right )\right )}{\left (x^{2}+4 x +4\right ) \left (-\frac {1}{x +2}\right )} \, dx} \\ &= \int { \frac {\left (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \left (-\frac {x}{2}-1-2 \cos \left (\ln \left (x +2\right )\right )-2 \sin \left (\ln \left (x +2\right )\right )\right )}{-\frac {x^{2}+4 x +4}{x +2}} \, dx}\\ &= \int {\left (\frac {x +2+4 \cos \left (\ln \left (x +2\right )\right )+4 \sin \left (\ln \left (x +2\right )\right )}{2 x +4}\right ) \, dx} \\ &= \frac {x}{2}+2 \sin \left (\ln \left (x +2\right )\right )-2 \cos \left (\ln \left (x +2\right )\right ) \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 (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \left (\cos \left (\ln \left (x +2\right )\right )\right )}{\left (x^{2}+4 x +4\right ) \left (-\frac {1}{x +2}\right )} \, dx} \\ &= - \int { \frac {\left (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \cos \left (\ln \left (x +2\right )\right )}{-\frac {x^{2}+4 x +4}{x +2}} \, dx}\\ &= - \int {\left (-\frac {\cos \left (\ln \left (x +2\right )\right )}{x +2}\right ) \, dx} \\ &= \sin \left (\ln \left (x +2\right )\right ) \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 (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \left (\sin \left (\ln \left (x +2\right )\right )\right )}{\left (x^{2}+4 x +4\right ) \left (-\frac {1}{x +2}\right )} \, dx} \\ &= \int { \frac {\left (-\left (x +2\right )^{2} y^{\prime \prime \prime }+1+\left (x^{2}+4 x +4\right ) y^{\prime \prime \prime }\right ) \sin \left (\ln \left (x +2\right )\right )}{-\frac {x^{2}+4 x +4}{x +2}} \, dx}\\ &= \int {\left (-\frac {\sin \left (\ln \left (x +2\right )\right )}{x +2}\right ) \, dx} \\ &= \cos \left (\ln \left (x +2\right )\right ) \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 {x}{2}+2 \sin \left (\ln \left (x +2\right )\right )-2 \cos \left (\ln \left (x +2\right )\right )\right ) \\ &+\left (\sin \left (\ln \left (x +2\right )\right )\right ) \left (-\frac {x \cos \left (\ln \left (x +2\right )\right )}{2}-\cos \left (\ln \left (x +2\right )\right )+\frac {x \sin \left (\ln \left (x +2\right )\right )}{2}+\sin \left (\ln \left (x +2\right )\right )-2\right ) \\ &+\left (\cos \left (\ln \left (x +2\right )\right )\right ) \left (\frac {x \cos \left (\ln \left (x +2\right )\right )}{2}+\cos \left (\ln \left (x +2\right )\right )+\frac {x \sin \left (\ln \left (x +2\right )\right )}{2}+\sin \left (\ln \left (x +2\right )\right )+2\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = x +1 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= \frac {\left (-c_{2} +c_{1} \right ) \left (x +2\right ) \cos \left (\ln \left (x +2\right )\right )}{2}+\frac {\left (c_{1} +c_{2} \right ) \left (x +2\right ) \sin \left (\ln \left (x +2\right )\right )}{2}+2 c_{1} -2 c_{2} +c_{3}\right ) + \left (x +1\right ) \\ \end{align*}

Summary

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

Verification of solutions

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

36.1.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +2\right )^{2} y^{\prime \prime \prime }+\left (x +2\right ) y^{\prime \prime }+y^{\prime }=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -((diff(_b(_a), _a))*_a+2*(diff(_b(_a), _a))+_b(_a)-1)/(_a+2)^2, _b(_ 
   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 
      <- LODE of Euler type successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
 

Solution by Maple

Time used: 0.016 (sec). Leaf size: 35

dsolve((x+2)^2*diff(y(x),x$3)+(x+2)*diff(y(x),x$2)+diff(y(x),x)=1,y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.202 (sec). Leaf size: 45

DSolve[(x+2)^2*y'''[x]+(x+2)*y''[x]+y'[x]==1,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to x+\frac {1}{2} (c_1-c_2) (x+2) \cos (\log (x+2))+\frac {1}{2} (c_1+c_2) (x+2) \sin (\log (x+2))+c_3 \]