2.1.3 problem 1 (iii)

Internal problem ID [18166]
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 : 1 (iii)
Date solved : Thursday, December 19, 2024 at 01:51:43 PM
CAS classification : [_quadrature]

Solve

\begin{align*} x^{\prime }&=\frac {1}{t^{2}+1} \end{align*}

With initial conditions

\begin{align*} x \left (1\right )&=0 \end{align*}

Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as

\begin{align*} x^{\prime } + q(t)x &= p(t) \end{align*}

Where here

\begin{align*} q(t) &=0\\ p(t) &=\frac {1}{t^{2}+1} \end{align*}

Hence the ode is

\begin{align*} x^{\prime } = \frac {1}{t^{2}+1} \end{align*}

The domain of \(q(t)=0\) is

\[ \{-\infty <t <\infty \} \]

And the point \(t_0 = 1\) is inside this domain. The domain of \(p(t)=\frac {1}{t^{2}+1}\) is

\[ \{-\infty <t <\infty \} \]

And the point \(t_0 = 1\) is also inside this domain. Hence solution exists and is unique.

Solved as first order quadrature ode

Time used: 0.105 (sec)

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

\begin{align*} \int {dx} &= \int {\frac {1}{t^{2}+1}\, dt}\\ x &= \arctan \left (t \right ) + c_1 \end{align*}

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} x = \arctan \left (t \right )-\frac {\pi }{4} \end{align*}

(a) Solution plot
\(x = \arctan \left (t \right )-\frac {\pi }{4}\)

(b) Slope field plot
\(x^{\prime } = \frac {1}{t^{2}+1}\)

Summary of solutions found

\begin{align*} x &= \arctan \left (t \right )-\frac {\pi }{4} \\ \end{align*}

Solved as first order Exact ode

Time used: 0.325 (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*} \mathop {\mathrm {d}x} &= \left (\frac {1}{t^{2}+1}\right )\mathop {\mathrm {d}t}\\ \left (-\frac {1}{t^{2}+1}\right ) \mathop {\mathrm {d}t} + \mathop {\mathrm {d}x} &= 0 \tag {2A} \end{align*}

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

\begin{align*} M(t,x) &= -\frac {1}{t^{2}+1}\\ N(t,x) &= 1 \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 (-\frac {1}{t^{2}+1}\right )\\ &= 0 \end{align*}

And

\begin{align*} \frac {\partial N}{\partial t} &= \frac {\partial }{\partial t} \left (1\right )\\ &= 0 \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 -\frac {1}{t^{2}+1}\mathop {\mathrm {d}t} \\ \tag{3} \phi &= -\arctan \left (t \right )+ 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} = 0+f'(x) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial x} = 1\). Therefore equation (4) becomes

\begin{equation} \tag{5} 1 = 0+f'(x) \end{equation}

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

\[ f'(x) = 1 \]

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

\begin{align*} \int f'(x) \mathop {\mathrm {d}x} &= \int \left ( 1\right ) \mathop {\mathrm {d}x} \\ f(x) &= x+ c_1 \\ \end{align*}

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

\[ \phi = -\arctan \left (t \right )+x+ c_1 \]

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

\[ c_1 = -\arctan \left (t \right )+x \]

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} -\arctan \left (t \right )+x = -\frac {\pi }{4} \end{align*}

Solving for \(x\) gives

\begin{align*} x &= \arctan \left (t \right )-\frac {\pi }{4} \\ \end{align*}

(a) Solution plot
\(x = \arctan \left (t \right )-\frac {\pi }{4}\)

(b) Slope field plot
\(x^{\prime } = \frac {1}{t^{2}+1}\)

Summary of solutions found

\begin{align*} x &= \arctan \left (t \right )-\frac {\pi }{4} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }=\frac {1}{t^{2}+1}, x \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int x^{\prime }d t =\int \frac {1}{t^{2}+1}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & x=\arctan \left (t \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\arctan \left (t \right )+\mathit {C1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (1\right )=0 \\ {} & {} & 0=\frac {\pi }{4}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =-\frac {\pi }{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =-\frac {\pi }{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\arctan \left (t \right )-\frac {\pi }{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\arctan \left (t \right )-\frac {\pi }{4} \end {array} \]

Maple trace

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

Maple dsolve solution

Solving time : 0.007 (sec)
Leaf size : 10

dsolve([diff(x(t),t) = 1/(t^2+1), 
       op([x(1) = 0])],x(t),singsol=all)
 

\[ x = \arctan \left (t \right )-\frac {\pi }{4} \]

Mathematica DSolve solution

Solving time : 0.005 (sec)
Leaf size : 13

DSolve[{D[x[t],t]==1/(1+t^2),{x[1]==0}}, 
       x[t],t,IncludeSingularSolutions->True]
 

\[ x(t)\to \arctan (t)-\frac {\pi }{4} \]