2.1.18 Problem 3 (vi)

Internal problem ID [18496]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 3. Solutions of first-order equations. Exercises at page 47
Problem number : 3 (vi)
Date solved : Saturday, February 22, 2025 at 09:17:54 PM
CAS classification : [_exact, _rational, [_1st_order, `_with_symmetry_[F(x),G(x)]`], [_Abel, `2nd type`, `class A`]]

Solve

\begin{align*} 2 t +3 x+\left (3 t -x\right ) x^{\prime }&=t^{2} \end{align*}

Solved as first order Exact ode

Time used: 0.180 (sec)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(t,x) \mathop {\mathrm {d}t}+ N(t,x) \mathop {\mathrm {d}x}=0 \tag {1A} \]

Therefore

\begin{align*} \left (3 t -x\right )\mathop {\mathrm {d}x} &= \left (t^{2}-2 t -3 x\right )\mathop {\mathrm {d}t}\\ \left (-t^{2}+2 t +3 x\right )\mathop {\mathrm {d}t} + \left (3 t -x\right )\mathop {\mathrm {d}x} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(t,x) &= -t^{2}+2 t +3 x\\ N(t,x) &= 3 t -x \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial x} = \frac {\partial N}{\partial t} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial x} &= \frac {\partial }{\partial x} \left (-t^{2}+2 t +3 x\right )\\ &= 3 \end{align*}

And

\begin{align*} \frac {\partial N}{\partial t} &= \frac {\partial }{\partial t} \left (3 t -x\right )\\ &= 3 \end{align*}

Since \(\frac {\partial M}{\partial x}= \frac {\partial N}{\partial t}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (t,x\right )\)

\begin{align*} \frac {\partial \phi }{\partial t } &= M\tag {1} \\ \frac {\partial \phi }{\partial x } &= N\tag {2} \end{align*}

Integrating (1) w.r.t. \(t\) gives

\begin{align*} \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int M\mathop {\mathrm {d}t} \\ \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int -t^{2}+2 t +3 x\mathop {\mathrm {d}t} \\ \tag{3} \phi &= -\frac {t \left (t^{2}-3 t -9 x \right )}{3}+ f(x) \\ \end{align*}

Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function of both \(t\) and \(x\). Taking derivative of equation (3) w.r.t \(x\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial x} = 3 t+f'(x) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial x} = 3 t -x\). Therefore equation (4) becomes

\begin{equation} \tag{5} 3 t -x = 3 t+f'(x) \end{equation}

Solving equation (5) for \( f'(x)\) gives

\[ f'(x) = -x \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int f'(x) \mathop {\mathrm {d}x} &= \int \left ( -x\right ) \mathop {\mathrm {d}x} \\ f(x) &= -\frac {x^{2}}{2}+ c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(x)\) into equation (3) gives \(\phi \)

\[ \phi = -\frac {t \left (t^{2}-3 t -9 x \right )}{3}-\frac {x^{2}}{2}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = -\frac {t \left (t^{2}-3 t -9 x \right )}{3}-\frac {x^{2}}{2} \]

Solving for \(x\) gives

\begin{align*} x &= 3 t -\frac {\sqrt {-6 t^{3}+99 t^{2}-18 c_1}}{3} \\ x &= 3 t +\frac {\sqrt {-6 t^{3}+99 t^{2}-18 c_1}}{3} \\ \end{align*}

Figure 2.31: Slope field \(2 t +3 x+\left (3 t -x\right ) x^{\prime } = t^{2}\)

Summary of solutions found

\begin{align*} x &= 3 t -\frac {\sqrt {-6 t^{3}+99 t^{2}-18 c_1}}{3} \\ x &= 3 t +\frac {\sqrt {-6 t^{3}+99 t^{2}-18 c_1}}{3} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 t +3 x+\left (3 t -x\right ) x^{\prime }=t^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \square & {} & \textrm {Check if ODE is exact}\hspace {3pt} \\ {} & \circ & \textrm {ODE is exact if the lhs is the total derivative of a}\hspace {3pt} C^{2}\hspace {3pt}\textrm {function}\hspace {3pt} \\ {} & {} & F^{\prime }\left (t , x\right )=0 \\ {} & \circ & \textrm {Compute derivative of lhs}\hspace {3pt} \\ {} & {} & F^{\prime }\left (t , x\right )+\left (\frac {\partial }{\partial x}F \left (t , x\right )\right ) x^{\prime }=0 \\ {} & \circ & \textrm {Evaluate derivatives}\hspace {3pt} \\ {} & {} & 3=3 \\ {} & \circ & \textrm {Condition met, ODE is exact}\hspace {3pt} \\ \bullet & {} & \textrm {Exact ODE implies solution will be of this form}\hspace {3pt} \\ {} & {} & \left [F \left (t , x\right )=\mathit {C1} , M \left (t , x\right )=F^{\prime }\left (t , x\right ), N \left (t , x\right )=\frac {\partial }{\partial x}F \left (t , x\right )\right ] \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} F \left (t , x\right )\hspace {3pt}\textrm {by integrating}\hspace {3pt} M \left (t , x\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} t \\ {} & {} & F \left (t , x\right )=\int \left (-t^{2}+2 t +3 x \right )d t +f_{1} \left (x \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & F \left (t , x\right )=-\frac {t^{3}}{3}+t^{2}+3 t x +f_{1} \left (x \right ) \\ \bullet & {} & \textrm {Take derivative of}\hspace {3pt} F \left (t , x\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} x \\ {} & {} & N \left (t , x\right )=\frac {\partial }{\partial x}F \left (t , x\right ) \\ \bullet & {} & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & 3 t -x =3 t +\frac {d}{d x}f_{1} \left (x \right ) \\ \bullet & {} & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}f_{1} \left (x \right ) \\ {} & {} & \frac {d}{d x}f_{1} \left (x \right )=-x \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} f_{1} \left (x \right ) \\ {} & {} & f_{1} \left (x \right )=-\frac {x^{2}}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} f_{1} \left (x \right )\hspace {3pt}\textrm {into equation for}\hspace {3pt} F \left (t , x\right ) \\ {} & {} & F \left (t , x\right )=-\frac {1}{3} t^{3}+t^{2}+3 t x -\frac {1}{2} x^{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} F \left (t , x\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & -\frac {1}{3} t^{3}+t^{2}+3 t x -\frac {1}{2} x^{2}=\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & \left \{x=3 t -\frac {\sqrt {-6 t^{3}+99 t^{2}-18 \mathit {C1}}}{3}, x=3 t +\frac {\sqrt {-6 t^{3}+99 t^{2}-18 \mathit {C1}}}{3}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
<- exact successful`
 

Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 51

dsolve(2*t+3*x(t)+(3*t-x(t))*diff(x(t),t) = t^2,x(t),singsol=all)
 

\begin{align*} x &= 3 t -\frac {\sqrt {-6 t^{3}+99 t^{2}+18 c_{1}}}{3} \\ x &= 3 t +\frac {\sqrt {-6 t^{3}+99 t^{2}+18 c_{1}}}{3} \\ \end{align*}

✓Mathematica DSolve solution

Solving time : 0.141 (sec)
Leaf size : 67

DSolve[{(2*t+3*x[t])+(3*t-x[t])*D[x[t],t]==t^2,{}},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)\to 3 t-i \sqrt {\frac {2 t^3}{3}-11 t^2-c_1} \\ x(t)\to 3 t+i \sqrt {\frac {2 t^3}{3}-11 t^2-c_1} \\ \end{align*}

✗Sympy solution

Solving time : 0.000 (sec)
Leaf size : 0

Python version: 3.13.1 (main, Dec  4 2024, 18:05:56) [GCC 14.2.1 20240910] 
Sympy version 1.13.3
 

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(-t**2 + 2*t + (3*t - x(t))*Derivative(x(t), t) + 3*x(t),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

Timed Out