2.1.56 Problem 56

Internal problem ID [8768]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 56
Date solved : Sunday, March 30, 2025 at 01:30:50 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order ode quadrature

Time used: 0.049 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }& =f\left(t\right)\end{array}$$

Integrating once gives

$$y^{\prime }=\int f\left(t\right)dt+c_1$$

Integrating again gives

$$y=\int (\int f\left(t\right)dt)dt+c_{1}x+c_2$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\int \int f\left(t\right)dtdt+t{c}_1+c_2\end{array}$$

Solved as second order linear constant coeff ode

Time used: 0.118 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }& =f\left(t\right)\end{array}$$

This is second order non-homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=f(t)$$

Where $A=1,B=0,C=0,f(t)=f\left(t\right)$. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=f(t)$. $y_h$ is the solution to

$$y^{\prime \prime }=0$$

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=0$$

Where in the above $A=1,B=0,C=0$. Let the solution be $y=e^{\lambda t}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{t\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda t}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=0$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{{\left(0\right)}^2-(4)\left(1\right)\left(0\right)}\\ & =0\end{array}$$

Hence this is the case of a double root ${\lambda }_{1,2}=0$. Therefore the solution is

$$\begin{array}{}\text{(1)}& y=c_{1}1+c_{2}t\end{array}$$

Therefore the homogeneous solution $y_h$ is

$$y_h=c_{2}t+c_1$$

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

$$\begin{array}{}\text{(1)}& y_p(t)=u_1{y}_1+u_2{y}_2\end{array}$$

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

$$\begin{array}{rl}{y}_1& =1\\ y_2& =t\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(t)}{aW(t)}\text{(3)}& u_2& =\int \frac{y_{1}f(t)}{aW(t)}\end{array}$$

Where $W(t)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}1& t\\ \frac{d}{dt}\left(1\right)& \frac{d}{dt}\left(t\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}1& t\\ 0& 1\end{array}\right|$$

Therefore

$$W=\left(1\right)\left(1\right)-\left(t\right)\left(0\right)$$

Which simplifies to

$$W=1$$

Which simplifies to

$$W=1$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{tf\left(t\right)}{1}dt$$

Which simplifies to

$$u_1=-\int tf\left(t\right)dt$$

Hence

$$u_1=-{\int }_0^t\alpha f\left(\alpha \right)d\alpha$$

And Eq. (3) becomes

$$u_2=\int \frac{f\left(t\right)}{1}dt$$

Which simplifies to

$$u_2=\int f\left(t\right)dt$$

Hence

$$u_2={\int }_0^{t}f\left(\alpha \right)d\alpha$$

Therefore the particular solution, from equation (1) is

$$y_p(t)=-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_{2}t+c_1)+(-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha )\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha +c_{2}t+c_1\end{array}$$

Solved as second order linear exact ode

Time used: 0.141 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }& =f\left(t\right)\end{array}$$

An ode of the form

$$\begin{array}{rl}p\left(t\right)y^{\prime \prime }+q\left(t\right)y^{\prime }+r\left(t\right)y& =s\left(t\right)\end{array}$$

is exact if

$$\begin{array}{}\text{(1)}& p^{″}(t)-q^{\prime }(t)+r(t)& =0\end{array}$$

For the given ode we have

$$\begin{array}{rl}p(x)& =1\\ q(x)& =0\\ r(x)& =0\\ s(x)& =f\left(t\right)\end{array}$$

Hence

$$\begin{array}{rl}{p}^{″}(x)& =0\\ q^{\prime }(x)& =0\end{array}$$

Therefore (1) becomes

$$\begin{array}{rl}0-\left(0\right)+\left(0\right)& =0\end{array}$$

Hence the ode is exact. Since we now know the ode is exact, it can be written as

$$\begin{array}{rl}{(p\left(t\right)y^{\prime }+(q\left(t\right)-p^{\prime }\left(t\right))y)}^{\prime }& =s(x)\end{array}$$

Integrating gives

$$\begin{array}{rl}p\left(t\right)y^{\prime }+(q\left(t\right)-p^{\prime }\left(t\right))y& =\int s\left(t\right)dt\end{array}$$

Substituting the above values for $p,q,r,s$ gives

$$\begin{array}{rl}{y}^{\prime }& =\int f\left(t\right)dt\end{array}$$

We now have a first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=\int f\left(t\right)dt+c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int dy& =\int \int f\left(t\right)dt+c_{1}dt\\ y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

$$\begin{array}{rl}y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

Solved as second order missing y ode

Time used: 0.080 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }& =f\left(t\right)\end{array}$$

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(t)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(t)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u}^{\prime }\left(t\right)-f\left(t\right)=0\end{array}$$

Which is now solved for $u(t)$ as first order ode.

Since the ode has the form $u^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int du& =\int f\left(t\right)dt\\ u& =\int f\left(t\right)dt+c_1\end{array}$$

$$\begin{array}{rl}u& =\int f\left(t\right)dt+c_1\end{array}$$

In summary, these are the solution found for $u(t)$

$$\begin{array}{rl}u& =\int f\left(t\right)dt+c_1\end{array}$$

For solution $u=\int f\left(t\right)dt+c_1$, since $u=y^{\prime }\left(t\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }\left(t\right)=\int f\left(t\right)dt+c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int dy& =\int \int f\left(t\right)dt+c_{1}dt\\ y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

$$\begin{array}{rl}y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\int (\int f\left(t\right)dt+c_1)dt+c_2\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.087 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }& =f\left(t\right)\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =0C& =0\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =y{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =0\end{array}$$

Equation (7) is now solved. After finding $z(t)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(t\right)e^{-\int \frac{B}{2A}dt}\end{array}$$

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 $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
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	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.21: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0--\mathrm{\infty }\\ & =\mathrm{\infty }\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $infinity$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=0$ is not a function of $t$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ 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

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =1\end{array}$$

Which simplifies to

$$y_1=1$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dt}}{y_1^2}dt$$

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dt\\ & =1\int \frac{1}{1}dt\\ & =1\left(t\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(1\right)+c_2\left(1\left(t\right)\right)\end{array}$$

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 $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=0$, and $y_p$ is a particular solution to the nonhomogeneous ODE $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=f(t)$. $y_h$ is the solution to

$$y^{\prime \prime }=0$$

The homogeneous solution is found using the Kovacic algorithm which results in

$$y_h=c_{2}t+c_1$$

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

$$\begin{array}{}\text{(1)}& y_p(t)=u_1{y}_1+u_2{y}_2\end{array}$$

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

$$\begin{array}{rl}{y}_1& =1\\ y_2& =t\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(t)}{aW(t)}\text{(3)}& u_2& =\int \frac{y_{1}f(t)}{aW(t)}\end{array}$$

Where $W(t)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}1& t\\ \frac{d}{dt}\left(1\right)& \frac{d}{dt}\left(t\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}1& t\\ 0& 1\end{array}\right|$$

Therefore

$$W=\left(1\right)\left(1\right)-\left(t\right)\left(0\right)$$

Which simplifies to

$$W=1$$

Which simplifies to

$$W=1$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{tf\left(t\right)}{1}dt$$

Which simplifies to

$$u_1=-\int tf\left(t\right)dt$$

Hence

$$u_1=-{\int }_0^t\alpha f\left(\alpha \right)d\alpha$$

And Eq. (3) becomes

$$u_2=\int \frac{f\left(t\right)}{1}dt$$

Which simplifies to

$$u_2=\int f\left(t\right)dt$$

Hence

$$u_2={\int }_0^{t}f\left(\alpha \right)d\alpha$$

Therefore the particular solution, from equation (1) is

$$y_p(t)=-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_{2}t+c_1)+(-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha )\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =-{\int }_0^t\alpha f\left(\alpha \right)d\alpha +t{\int }_0^{t}f\left(\alpha \right)d\alpha +c_{2}t+c_1\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 15

ode:=diff(diff(y(t),t),t) = f(t); 
dsolve(ode,y(t), singsol=all);
 

$$y=\int \int f\left(t\right)dtdt+c_{1}t+c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=f\left(t\right)\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{0\pm \left(\sqrt{0}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=0\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(t\right)=1\\ \bullet & & \text{Repeated root, multiply}{y}_1\left(t\right)\text{by}t\text{to ensure linear independence}\\ & & y_2\left(t\right)=t\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)+y_p\left(t\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}+\mathit{C}\mathit{2}t+y_p\left(t\right)\\ ◻& & \text{Find a particular solution}{y}_p\left(t\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{y}_p\text{here}f\left(t\right)\text{is the forcing function}\\ & & [y_p\left(t\right)=-y_1\left(t\right)\int \frac{y_2\left(t\right)f\left(t\right)}{W(y_1\left(t\right),y_2\left(t\right))}dt+y_2\left(t\right)\int \frac{y_1\left(t\right)f\left(t\right)}{W(y_1\left(t\right),y_2\left(t\right))}dt,f\left(t\right)=f\left(t\right)]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(y_1\left(t\right),y_2\left(t\right))=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(y_1\left(t\right),y_2\left(t\right))=1\\ & \circ & \text{Substitute functions into equation for}{y}_p\left(t\right)\\ & & y_p\left(t\right)=-\int tf\left(t\right)dt+t\int f\left(t\right)dt\\ & \circ & \text{Compute integrals}\\ & & y_p\left(t\right)=-\int tf\left(t\right)dt+t\int f\left(t\right)dt\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}+\mathit{C}\mathit{2}t-\int tf\left(t\right)dt+t\int f\left(t\right)dt\end{array}$$

✓ Mathematica. Time used: 0.011 (sec). Leaf size: 30

ode=D[y[t],{t,2}]==f[t]; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to {\int }_1^t{\int }_1^{K[2]}f(K[1])dK[1]dK[2]+c_{2}t+c_1$$

✓ Sympy. Time used: 0.312 (sec). Leaf size: 19

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-f(t) + Derivative(y(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_1+t(C_2+\int f\left(t\right)dt)-\int tf\left(t\right)dt$$