2.5.4 Problem 5

Internal problem ID [18491]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 32. Problems at page 89
Problem number : 5
Date solved : Monday, March 31, 2025 at 05:37:26 PM
CAS classification : [[_homogeneous, `class G`]]

Solved using first_order_ode_isobaric

Time used: 0.358 (sec)

Solve

$$\begin{array}{rl}\sqrt{t^2+T}& =T^{\prime }\end{array}$$

Solving for $T^{\prime }$ gives

$$\begin{array}{}\text{(1)}& T^{\prime }& =\sqrt{t^2+T}\end{array}$$

Each of the above ode’s is now solved An ode $T^{\prime }=f(t,T)$ is isobaric if

$$\begin{array}{}\text{(1)}& f(tt,t^{m}T)=t^{m-1}f(t,T)\end{array}$$

Where here

$$\begin{array}{}\text{(2)}& f(t,T)=\sqrt{t^2+T}\end{array}$$

$m$ is the order of isobaric. Substituting (2) into (1) and solving for $m$ gives

$$m=2$$

Since the ode is isobaric of order $m=2$, then the substitution

$$\begin{array}{rl}T& =u{t}^m\\ & =u{t}^2\end{array}$$

Converts the ODE to a separable in $u\left(t\right)$. Performing this substitution gives

$$2tu\left(t\right)+t^2{u}^{\prime }\left(t\right)=\sqrt{t^2+t^{2}u\left(t\right)}$$

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }\left(t\right)=-\frac{2u\left(t\right)-\sqrt{1+u\left(t\right)}}{t}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }\left(t\right)& =-\frac{2u\left(t\right)-\sqrt{1+u\left(t\right)}}{t}\\ & =f(t)g(u)\end{array}$$

Where

$$\begin{array}{rl}f(t)& =\frac{1}{t}\\ g(u)& =-2u+\sqrt{u+1}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(t)dt\\ \int \frac{1}{-2u+\sqrt{u+1}}du& =\int \frac{1}{t}dt\end{array}$$

$$-\frac{\ln (2u\left(t\right)-\sqrt{1+u\left(t\right)})}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{1+u\left(t\right)}-1)\sqrt{17}}{17}\right)}{17}=\ln \left(t\right)+c_1$$

We now need to find the singular solutions, these are found by finding for what values $g(u)$ is zero, since we had to divide by this above. Solving $g(u)=0$ or

$$-2u+\sqrt{u+1}=0$$

for $u\left(t\right)$ gives

$$\begin{array}{rl}u\left(t\right)& =\frac{1}{8}+\frac{\sqrt{17}}{8}\end{array}$$

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

$$\begin{array}{rl}-\frac{\ln (2u\left(t\right)-\sqrt{1+u\left(t\right)})}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{1+u\left(t\right)}-1)\sqrt{17}}{17}\right)}{17}& =\ln \left(t\right)+c_1\\ u\left(t\right)& =\frac{1}{8}+\frac{\sqrt{17}}{8}\end{array}$$

Converting $-\frac{\ln (2u\left(t\right)-\sqrt{1+u\left(t\right)})}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{1+u\left(t\right)}-1)\sqrt{17}}{17}\right)}{17}=\ln \left(t\right)+c_1$ back to $T$ gives

$$\begin{array}{r}-\frac{\ln (\frac{2T}{t^2}-\sqrt{1+\frac{T}{t^2}})}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{1+\frac{T}{t^2}}-1)\sqrt{17}}{17}\right)}{17}=\ln \left(t\right)+c_1\end{array}$$

Converting $u\left(t\right)=\frac{1}{8}+\frac{\sqrt{17}}{8}$ back to $T$ gives

$$\begin{array}{r}\frac{T}{t^2}=\frac{1}{8}+\frac{\sqrt{17}}{8}\end{array}$$

Solving for $T$ gives

$$\begin{array}{rl}-\frac{\ln (\frac{2T}{t^2}-\sqrt{1+\frac{T}{t^2}})}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{1+\frac{T}{t^2}}-1)\sqrt{17}}{17}\right)}{17}& =\ln \left(t\right)+c_1\\ T& =\frac{(1+\sqrt{17})t^2}{8}\end{array}$$

Which simplifies to

$$\begin{array}{rl}-\frac{\ln \left(\frac{-\sqrt{\frac{t^2+T}{t^2}}{t}^2+2T}{t^2}\right)}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{\frac{t^2+T}{t^2}}-1)\sqrt{17}}{17}\right)}{17}& =\ln \left(t\right)+c_1\\ T& =\frac{(1+\sqrt{17})t^2}{8}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}-\frac{\ln \left(\frac{-\sqrt{\frac{t^2+T}{t^2}}{t}^2+2T}{t^2}\right)}{2}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{\frac{t^2+T}{t^2}}-1)\sqrt{17}}{17}\right)}{17}& =\ln \left(t\right)+c_1\\ T& =\frac{(1+\sqrt{17})t^2}{8}\end{array}$$

Solved using first_order_ode_LIE

Time used: 1.517 (sec)

Solve

$$\begin{array}{rl}\sqrt{t^2+T}& =T^{\prime }\end{array}$$

Writing the ode as

$$\begin{array}{rl}{T}^{\prime }& =\sqrt{t^2+T}\\ T^{\prime }& =\omega (t,T)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

$$\begin{array}{}\text{(A)}& {\eta }_t+\omega ({\eta }_T-{\xi }_t)-{\omega }^2{\xi }_T-{\omega }_t\xi -{\omega }_T\eta =0\end{array}$$

To determine $\xi ,\eta$ then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

$$\begin{array}{}\text{(1E)}& \xi & =T{a}_3+t{a}_2+a_1\text{(2E)}& \eta & =T{b}_3+t{b}_2+b_1\end{array}$$

Where the unknown coefficients are

$$\{a_1,a_2,a_3,b_1,b_2,b_3\}$$

Substituting equations (1E,2E) and $\omega$ into (A) gives

$$\begin{array}{}\text{(5E)}& b_2+\sqrt{t^2+T}(b_3-a_2)-(t^2+T)a_3-\frac{t(T{a}_3+t{a}_2+a_1)}{\sqrt{t^2+T}}-\frac{T{b}_3+t{b}_2+b_1}{2\sqrt{t^2+T}}=0\end{array}$$

Putting the above in normal form gives

$$-\frac{2\sqrt{t^2+T}{t}^2{a}_3+2\sqrt{t^2+T}T{a}_3+2Tt{a}_3+4t^2{a}_2-2t^2{b}_3-2b_2\sqrt{t^2+T}+2T{a}_2-T{b}_3+2t{a}_1+t{b}_2+b_1}{2\sqrt{t^2+T}}=0$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& -2\sqrt{t^2+T}{t}^2{a}_3-2\sqrt{t^2+T}T{a}_3-2Tt{a}_3-4t^2{a}_2+2t^2{b}_3+2b_2\sqrt{t^2+T}-2T{a}_2+T{b}_3-2t{a}_1-t{b}_2-b_1=0\end{array}$$

Simplifying the above gives

$$\begin{array}{}\text{(6E)}& -2\sqrt{t^2+T}{t}^2{a}_3-2(t^2+T)a_2+2(t^2+T)b_3-2\sqrt{t^2+T}T{a}_3-2Tt{a}_3-2t^2{a}_2+2b_2\sqrt{t^2+T}-T{b}_3-2t{a}_1-t{b}_2-b_1=0\end{array}$$

Since the PDE has radicals, simplifying gives

$$-2\sqrt{t^2+T}{t}^2{a}_3-2\sqrt{t^2+T}T{a}_3-2Tt{a}_3-4t^2{a}_2+2t^2{b}_3+2b_2\sqrt{t^2+T}-2T{a}_2+T{b}_3-2t{a}_1-t{b}_2-b_1=0$$

Looking at the above PDE shows the following are all the terms with $\{T,t\}$ in them.

$$\{T,t,\sqrt{t^2+T}\}$$

The following substitution is now made to be able to collect on all terms with $\{T,t\}$ in them

$$\{T=v_1,t=v_2,\sqrt{t^2+T}=v_3\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& -2v_3{v}_2^2{a}_3-4v_2^2{a}_2-2v_1{v}_2{a}_3-2v_3{v}_1{a}_3+2v_2^2{b}_3-2v_2{a}_1-2v_1{a}_2-v_2{b}_2+2b_2{v}_3+v_1{b}_3-b_1=0\end{array}$$

Collecting the above on the terms $v_i$ introduced, and these are

$$\{v_1,v_2,v_3\}$$

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& -2v_1{v}_2{a}_3-2v_3{v}_1{a}_3+(-2a_2+b_3)v_1-2v_3{v}_2^2{a}_3+(-4a_2+2b_3)v_2^2+(-2a_1-b_2)v_2+2b_2{v}_3-b_1=0\end{array}$$

Setting each coefficients in (8E) to zero gives the following equations to solve

$$\begin{array}{rl}-2a_3& =0\\ -b_1& =0\\ 2b_2& =0\\ -2a_1-b_2& =0\\ -4a_2+2b_3& =0\\ -2a_2+b_3& =0\end{array}$$

Solving the above equations for the unknowns gives

$$\begin{array}{rl}{a}_1& =0\\ a_2& =a_2\\ a_3& =0\\ b_1& =0\\ b_2& =0\\ b_3& =2a_2\end{array}$$

Substituting the above solution in the anstaz (1E,2E) (using $1$ as arbitrary value for any unknown in the RHS) gives

$$\begin{array}{rl}\xi & =t\\ \eta & =2T\end{array}$$

Shifting is now applied to make $\xi =0$ in order to simplify the rest of the computation

$$\begin{array}{rl}\eta & =\eta -\omega (t,T)\xi \\ & =2T-\left(\sqrt{t^2+T}\right)\left(t\right)\\ & =-t\sqrt{t^2+T}+2T\\ \xi & =0\end{array}$$

The next step is to determine the canonical coordinates $R,S$. The canonical coordinates map $(t,T)\to (R,S)$ where $(R,S)$ are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

$$\begin{array}{}\text{(1)}& \frac{dt}{\xi }& =\frac{dT}{\eta }=dS\end{array}$$

The above comes from the requirements that $(\xi \frac{\partial }{\partial t}+\eta \frac{\partial }{\partial T})S(t,T)=1$. Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable $R$ in the canonical coordinates, where $S(R)$. Since $\xi =0$ then in this special case

$$\begin{array}{r}R=t\end{array}$$

$S$ is found from

$$\begin{array}{rl}S& =\int \frac{1}{\eta }dy\\ & =\int \frac{1}{-t\sqrt{t^2+T}+2T}dy\end{array}$$

Which results in

$$\begin{array}{rl}S& =-\frac{\ln (t\sqrt{t^2+T}+2T)}{4}-\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{t^2+T}+t)\sqrt{17}}{17t}\right)}{34}+\frac{\ln (-t\sqrt{t^2+T}+2T)}{4}-\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(-t+4\sqrt{t^2+T})\sqrt{17}}{17t}\right)}{34}+\frac{\ln (-t^4-T{t}^2+4T^2)}{4}-\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(-t^2+8T)\sqrt{17}}{17t^2}\right)}{34}\end{array}$$

Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_t+\omega (t,T)S_T}{R_t+\omega (t,T)R_T}\end{array}$$

Where in the above $R_t,R_T,S_t,S_T$ are all partial derivatives and $\omega (t,T)$ is the right hand side of the original ode given by

$$\begin{array}{rl}\omega (t,T)& =\sqrt{t^2+T}\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_t& =1\\ R_T& =0\\ S_t& =\frac{t^6+2T{t}^4-3T^2{t}^2-4T^3}{{(t\sqrt{t^2+T}-2T)}^2(t\sqrt{t^2+T}+2T)\sqrt{t^2+T}}\\ S_T& =\frac{(2\sqrt{t^2+T}+t)T+t^3}{(-t^4-T{t}^2+4T^2)\sqrt{t^2+T}}\end{array}$$

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =0\end{array}$$

We now need to express the RHS as function of $R$ only. This is done by solving for $t,T$ in terms of $R,S$ from the result obtained earlier and simplifying. This gives

$$\begin{array}{rl}\frac{dS}{dR}& =0\end{array}$$

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates $R,S$.

Since the ode has the form $\frac{d}{dR}S\left(R\right)=f(R)$, then we only need to integrate $f(R)$.

$$\begin{array}{rl}\int dS& =\int 0dR+c_3\\ S\left(R\right)& =c_3\end{array}$$

To complete the solution, we just need to transform the above back to $t,T$ coordinates. This results in

$$\begin{array}{r}-\frac{\ln (t\sqrt{t^2+T}+2T)}{4}-\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(4\sqrt{t^2+T}+t)\sqrt{17}}{17t}\right)}{34}+\frac{\ln (-t\sqrt{t^2+T}+2T)}{4}+\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(t-4\sqrt{t^2+T})\sqrt{17}}{17t}\right)}{34}+\frac{\ln (-t^4-T{t}^2+4T^2)}{4}-\frac{\sqrt{17}\mathrm{arctanh}\left(\frac{(-t^2+8T)\sqrt{17}}{17t^2}\right)}{34}=c_3\end{array}$$

Which simplifies to

$$\begin{array}{rl}\frac{\ln (-t^4-T{t}^2+4T^2)}{4}+\frac{\ln (-t\sqrt{t^2+T}+2T)}{4}-\frac{\ln (t\sqrt{t^2+T}+2T)}{4}+\frac{(-2\mathrm{arctanh}\left(\frac{(4\sqrt{t^2+T}+t)\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t-4\sqrt{t^2+T})\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t^2-8T)\sqrt{17}}{17t^2}\right))\sqrt{17}}{68}& =c_3\end{array}$$

Summary of solutions found

$$\begin{array}{rl}\frac{\ln (-t^4-T{t}^2+4T^2)}{4}+\frac{\ln (-t\sqrt{t^2+T}+2T)}{4}-\frac{\ln (t\sqrt{t^2+T}+2T)}{4}+\frac{(-2\mathrm{arctanh}\left(\frac{(4\sqrt{t^2+T}+t)\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t-4\sqrt{t^2+T})\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t^2-8T)\sqrt{17}}{17t^2}\right))\sqrt{17}}{68}& =c_3\end{array}$$

✓ Maple. Time used: 0.012 (sec). Leaf size: 136

ode:=(t^2+T(t))^(1/2) = diff(T(t),t); 
dsolve(ode,T(t), singsol=all);
 

$$17\ln (-t^4-T{t}^2+4T^2)+17\ln (-\sqrt{t^2+T}t+2T)-17\ln (\sqrt{t^2+T}t+2T)+(-2\mathrm{arctanh}\left(\frac{(4\sqrt{t^2+T}+t)\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t-4\sqrt{t^2+T})\sqrt{17}}{17t}\right)+2\mathrm{arctanh}\left(\frac{(t^2-8T)\sqrt{17}}{17t^2}\right))\sqrt{17}-c_1=0$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous G 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful 
<- homogeneous successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \sqrt{t^2+T\left(t\right)}=\frac{d}{dt}T\left(t\right)\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}T\left(t\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dt}T\left(t\right)=\sqrt{t^2+T\left(t\right)}\end{array}$$

✓ Mathematica. Time used: 0.277 (sec). Leaf size: 135

ode=Sqrt[t^2+T[t]]==D[T[t],t]; 
ic={}; 
DSolve[{ode,ic},T[t],t,IncludeSingularSolutions->True]
 

$$\text{Solve}[\frac{1}{34}(-34\log (\sqrt{t^2+T(t)}-t)-(\sqrt{17}-17)\log (2(\sqrt{17}-4)t\sqrt{t^2+T(t)}-2(\sqrt{17}-4)t^2-(\sqrt{17}-3)T(t))+(17+\sqrt{17})\log (2(4+\sqrt{17})t\sqrt{t^2+T(t)}-2(4+\sqrt{17})t^2-(3+\sqrt{17})T(t)))=c_1,T(t)]$$

✗ Sympy

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

NotImplementedError : The given ODE -sqrt(t**2 + T(t)) + Derivative(T(t), t) cannot be solved by the factorable group method