Internal problem ID [14635]
Internal file name [OUTPUT/14315_Thursday_March_28_2024_03_35_16_AM_83276714/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.4, page 163
Problem number: 65.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (\sin \left (t \right )-t \cos \left (t \right )\right ) y^{\prime \prime }-t \sin \left (t \right ) y^{\prime }+\sin \left (t \right ) y=t} \] With initial conditions \begin {align*} \left [y \left (\frac {\pi }{4}\right ) = 0, y^{\prime }\left (\frac {\pi }{4}\right ) = 0\right ] \end {align*}
This is second order non-homogeneous ODE. In standard form the ODE is \[ A y''(t) + B y'(t) + C y(t) = f(t) \] Where \(A=\sin \left (t \right )-t \cos \left (t \right ), B=-t \sin \left (t \right ), C=\sin \left (t \right ), f(t)=t\). 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 non-homogeneous ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). Solving for \(y_h\) from \[ \left (\sin \left (t \right )-t \cos \left (t \right )\right ) y^{\prime \prime }-t \sin \left (t \right ) y^{\prime }+\sin \left (t \right ) y = 0 \] In normal form the ode \begin {align*} \left (\sin \left (t \right )-t \cos \left (t \right )\right ) y^{\prime \prime }-t \sin \left (t \right ) y^{\prime }+\sin \left (t \right ) 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 {t \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}\\ q \left (t \right )&=-\frac {\sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )} \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 {n \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}-\frac {\sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}&=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 (t \right )+\left (\frac {2}{t}+\frac {t \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}\right ) v^{\prime }\left (t \right )&=0 \\ v^{\prime \prime }\left (t \right )+\left (\frac {2}{t}+\frac {t \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}\right ) v^{\prime }\left (t \right )&=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 )+\left (\frac {2}{t}+\frac {t \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )}\right ) u \left (t \right ) = 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 {u \left (2 t \cos \left (t \right )-2 \sin \left (t \right )+\sin \left (t \right ) t^{2}\right )}{t \left (-\sin \left (t \right )+t \cos \left (t \right )\right )} \end {align*}
Where \(f(t)=-\frac {2 t \cos \left (t \right )-2 \sin \left (t \right )+\sin \left (t \right ) t^{2}}{t \left (-\sin \left (t \right )+t \cos \left (t \right )\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2 t \cos \left (t \right )-2 \sin \left (t \right )+\sin \left (t \right ) t^{2}}{t \left (-\sin \left (t \right )+t \cos \left (t \right )\right )} \,d t\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2 t \cos \left (t \right )-2 \sin \left (t \right )+\sin \left (t \right ) t^{2}}{t \left (-\sin \left (t \right )+t \cos \left (t \right )\right )} \,d t}\\ \ln \left (u \right )&=-2 \ln \left (t \right )-\ln \left (1+\tan \left (\frac {t}{2}\right )^{2}\right )+\ln \left (t \tan \left (\frac {t}{2}\right )^{2}-t +2 \tan \left (\frac {t}{2}\right )\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (t \right )-\ln \left (1+\tan \left (\frac {t}{2}\right )^{2}\right )+\ln \left (t \tan \left (\frac {t}{2}\right )^{2}-t +2 \tan \left (\frac {t}{2}\right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (t \right )-\ln \left (1+\tan \left (\frac {t}{2}\right )^{2}\right )+\ln \left (t \tan \left (\frac {t}{2}\right )^{2}-t +2 \tan \left (\frac {t}{2}\right )\right )} \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} \sin \left (t \right )}{t}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (t \right ) t^{n}\\ &= \left (-\frac {c_{1} \sin \left (t \right )}{t}+c_{2} \right ) t\\ &= -\sin \left (t \right ) c_{1} +c_{2} t\\ \end {align*}
Now the particular solution to this ODE is found \[ \left (\sin \left (t \right )-t \cos \left (t \right )\right ) y^{\prime \prime }-t \sin \left (t \right ) y^{\prime }+\sin \left (t \right ) y = t \] 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 \\ y_2 &= \sin \left (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 \[ W = \begin {vmatrix} t & \sin \left (t \right ) \\ \frac {d}{dt}\left (t\right ) & \frac {d}{dt}\left (\sin \left (t \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} t & \sin \left (t \right ) \\ 1 & \cos \left (t \right ) \end {vmatrix} \] Therefore \[ W = \left (t\right )\left (\cos \left (t \right )\right ) - \left (\sin \left (t \right )\right )\left (1\right ) \] Which simplifies to \[ W = -\sin \left (t \right )+t \cos \left (t \right ) \] Which simplifies to \[ W = -\sin \left (t \right )+t \cos \left (t \right ) \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {t \sin \left (t \right )}{\left (\sin \left (t \right )-t \cos \left (t \right )\right ) \left (-\sin \left (t \right )+t \cos \left (t \right )\right )}\,dt \] Which simplifies to \[ u_1 = - \int -\frac {\sin \left (t \right ) t}{\left (-\sin \left (t \right )+t \cos \left (t \right )\right )^{2}}d t \] Hence \[ u_1 = \frac {1}{-\sin \left (t \right )+t \cos \left (t \right )} \] And Eq. (3) becomes \[ u_2 = \int \frac {t^{2}}{\left (\sin \left (t \right )-t \cos \left (t \right )\right ) \left (-\sin \left (t \right )+t \cos \left (t \right )\right )}\,dt \] Which simplifies to \[ u_2 = \int -\frac {t^{2}}{\left (-\sin \left (t \right )+t \cos \left (t \right )\right )^{2}}d t \] Hence \[ u_2 = \frac {1-\tan \left (\frac {t}{2}\right )^{2}+2 \tan \left (\frac {t}{2}\right ) t}{t \tan \left (\frac {t}{2}\right )^{2}-t +2 \tan \left (\frac {t}{2}\right )} \] Which simplifies to \begin{align*} u_1 &= \frac {1}{-\sin \left (t \right )+t \cos \left (t \right )} \\ u_2 &= \frac {-t \sin \left (t \right )-\cos \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(t) = \frac {t}{-\sin \left (t \right )+t \cos \left (t \right )}+\frac {\left (-t \sin \left (t \right )-\cos \left (t \right )\right ) \sin \left (t \right )}{-\sin \left (t \right )+t \cos \left (t \right )} \] Which simplifies to \[ y_p(t) = \cos \left (t \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (-\frac {c_{1} \sin \left (t \right )}{t}+c_{2} \right ) t\right ) + \left (\cos \left (t \right )\right ) \\ &= \cos \left (t \right )+\left (-\frac {c_{1} \sin \left (t \right )}{t}+c_{2} \right ) t \\ \end{align*} Which simplifies to \[ y = -\sin \left (t \right ) c_{1} +c_{2} t +\cos \left (t \right ) \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\sin \left (t \right ) c_{1} +c_{2} t +\cos \left (t \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 \(t = \frac {\pi }{4}\) in the above gives \begin {align*} 0 = \frac {\left (2-2 c_{1} \right ) \sqrt {2}}{4}+\frac {\pi c_{2}}{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} \cos \left (t \right )+c_{2} -\sin \left (t \right ) \end {align*}
substituting \(y^{\prime } = 0\) and \(t = \frac {\pi }{4}\) in the above gives \begin {align*} 0 = \frac {\left (-1-c_{1} \right ) \sqrt {2}}{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}&=-\frac {\pi +4}{-4+\pi }\\ c_{2}&=-\frac {4 \sqrt {2}}{-4+\pi } \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {\pi \cos \left (t \right )+\pi \sin \left (t \right )-4 \sqrt {2}\, t -4 \cos \left (t \right )+4 \sin \left (t \right )}{-4+\pi } \end {align*}
Which simplifies to \[ y = \frac {-4 \sqrt {2}\, t +\left (-4+\pi \right ) \cos \left (t \right )+\sin \left (t \right ) \left (\pi +4\right )}{-4+\pi } \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-4 \sqrt {2}\, t +\left (-4+\pi \right ) \cos \left (t \right )+\sin \left (t \right ) \left (\pi +4\right )}{-4+\pi } \\ \end{align*}
Verification of solutions
\[ y = \frac {-4 \sqrt {2}\, t +\left (-4+\pi \right ) \cos \left (t \right )+\sin \left (t \right ) \left (\pi +4\right )}{-4+\pi } \] Verified OK.
Maple trace
`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)] trying symmetries linear in x and y(x) Try integration with the canonical coordinates of the symmetry [0, t] -> Calling odsolve with the ODE`, diff(_b(_a), _a) = (sin(_a)*_a^2*_b(_a)+2*cos(_a)*_a*_b(_a)-2*sin(_a)*_b(_a)+_a)/((-cos(_a)*_a+sin Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- differential order: 2; canonical coordinates successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 29
dsolve([(sin(t)-t*cos(t))*diff(y(t),t$2)-t*sin(t)*diff(y(t),t)+sin(t)*y(t)=t,y(1/4*Pi) = 0, D(y)(1/4*Pi) = 0],y(t), singsol=all)
\[ y \left (t \right ) = \frac {-4 \sqrt {2}\, t +\left (-4+\pi \right ) \cos \left (t \right )+\left (\pi +4\right ) \sin \left (t \right )}{-4+\pi } \]
✓ Solution by Mathematica
Time used: 1.047 (sec). Leaf size: 32
DSolve[{(Sin[t]-t*Cos[t])*y''[t]-t*Sin[t]*y'[t]+Sin[t]*y[t]==t,{y[Pi/4]==0,y'[Pi/4]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {-4 \sqrt {2} t+(4+\pi ) \sin (t)+(\pi -4) \cos (t)}{\pi -4} \]