2.1.4 problem 4

Internal problem ID [8664]
Book : First order enumerated odes
Section : section 1
Problem number : 4
Date solved : Tuesday, December 17, 2024 at 12:57:15 PM
CAS classification : [_quadrature]

Solve

\begin{align*} y^{\prime }&=1 \end{align*}

Solved as first order quadrature ode

Time used: 0.032 (sec)

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

\begin{align*} \int {dy} &= \int {1\, dx}\\ y &= x + c_1 \end{align*}

Figure 2.6: Slope field plot
\(y^{\prime } = 1\)

Summary of solutions found

\begin{align*} y &= x +c_1 \\ \end{align*}

Solved as first order homogeneous class D2 ode

Time used: 0.173 (sec)

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = 1 \end{align*}

Which is now solved The ode \(u^{\prime }\left (x \right ) = -\frac {u \left (x \right )-1}{x}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )-1}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -u +1 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{-u +1}\,du} &= \int { \frac {1}{x} \,dx}\\ -\ln \left (u \left (x \right )-1\right )&=\ln \left (x \right )+c_1 \end{align*}

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 \(-u +1=0\) for \(u \left (x \right )\) gives

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

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{align*} -\ln \left (u \left (x \right )-1\right ) = \ln \left (x \right )+c_1\\ u \left (x \right ) = 1 \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= 1 \\ u \left (x \right ) &= \frac {\left (x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1}}{x} \\ \end{align*}

Converting \(u \left (x \right ) = 1\) back to \(y\) gives

\begin{align*} y = x \end{align*}

Converting \(u \left (x \right ) = \frac {\left (x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1}}{x}\) back to \(y\) gives

\begin{align*} y = \left (x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1} \end{align*}

Figure 2.7: Slope field plot
\(y^{\prime } = 1\)

Summary of solutions found

\begin{align*} y &= x \\ y &= \left (x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1} \\ \end{align*}

Solved as first order Exact ode

Time used: 0.056 (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

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

Therefore

\begin{align*} \left (1\right )\mathop {\mathrm {d}y} &= \mathop {\mathrm {d}x}\\ - \mathop {\mathrm {d}x} + \left (1\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

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

\begin{align*} M(x,y) &= -1\\ N(x,y) &= 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

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

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-1\right )\\ &= 0 \end{align*}

And

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

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

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

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

\begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -1\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -x+ f(y) \\ \end{align*}

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

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

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

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

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

\[ f'(y) = 1 \]

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

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

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

\[ \phi = -x +y+ c_1 \]

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

\[ c_1 = -x +y \]

Solving for \(y\) gives

\begin{align*} y &= x +c_1 \\ \end{align*}

Figure 2.8: Slope field plot
\(y^{\prime } = 1\)

Summary of solutions found

\begin{align*} y &= x +c_1 \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y \left (x \right )=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=x +\mathit {C1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 7

dsolve(diff(y(x),x) = 1, 
       y(x),singsol=all)
 

\[ y = x +c_{1} \]

Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 9

DSolve[{D[y[x],x]==1,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to x+c_1 \]