1.8 problem 8

Internal problem ID [7700]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 8
Date solved : Monday, October 21, 2024 at 03:57:25 PM
CAS classification : [_quadrature]

Solve

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

1.8.1 Solved as first order quadrature ode

Time used: 0.049 (sec)

Integrating gives

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

Figure 15: Slope field plot
\(y^{\prime } = y\)

1.8.2 Solved as first order homogeneous class D2 ode

Time used: 0.135 (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 ) = u \left (x \right ) x \end{align*}

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

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

Where

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

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u}\,du} &= \int { \frac {x -1}{x} \,dx}\\ \ln \left (u \left (x \right )\right )&=x +\ln \left (\frac {1}{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 \(u=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \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 )\right ) = x +\ln \left (\frac {1}{x}\right )+c_1\\ u \left (x \right ) = 0 \end{align*}

Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=\frac {{\mathrm e}^{x +c_1}}{x} \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

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

\begin{align*} y = {\mathrm e}^{x +c_1} \end{align*}

Figure 16: Slope field plot
\(y^{\prime } = y\)

1.8.3 Solved as first order Exact ode

Time used: 0.092 (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*} \mathop {\mathrm {d}y} &= \left (y\right )\mathop {\mathrm {d}x}\\ \left (-y\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

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

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

And

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

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

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

Since \(A\) does not depend on \(y\), then it can be used to find an integrating factor. The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{ \int A \mathop {\mathrm {d}x} } \\ &= e^{\int -1\mathop {\mathrm {d}x} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-x } \\ &= {\mathrm e}^{-x} \end{align*}

\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) for now so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= {\mathrm e}^{-x}\left (-y\right ) \\ &= -y \,{\mathrm e}^{-x} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{-x}\left (1\right ) \\ &= {\mathrm e}^{-x} \end{align*}

Now a modified ODE is ontained from the original ODE, which is exact and can be solved. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-y \,{\mathrm e}^{-x}\right ) + \left ({\mathrm e}^{-x}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

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

Integrating (2) w.r.t. \(y\) gives

\begin{align*} \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int \overline {N}\mathop {\mathrm {d}y} \\ \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int {\mathrm e}^{-x}\mathop {\mathrm {d}y} \\ \tag{3} \phi &= y \,{\mathrm e}^{-x}+ f(x) \\ \end{align*}

Where \(f(x)\) 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 \(x\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial x} = -y \,{\mathrm e}^{-x}+f'(x) \end{equation}

But equation (1) says that \(\frac {\partial \phi }{\partial x} = -y \,{\mathrm e}^{-x}\). Therefore equation (4) becomes

\begin{equation} \tag{5} -y \,{\mathrm e}^{-x} = -y \,{\mathrm e}^{-x}+f'(x) \end{equation}

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

\[ f'(x) = 0 \]

Therefore

\[ f(x) = c_1 \]

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

\[ \phi = y \,{\mathrm e}^{-x}+ c_1 \]

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

\[ c_1 = y \,{\mathrm e}^{-x} \]

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y = c_1 \,{\mathrm e}^{x} \end{align*}

Figure 17: Slope field plot
\(y^{\prime } = y\)

1.8.4 Maple step by step solution

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

1.8.5 Maple trace

Methods for first order ODEs:
 

1.8.6 Maple dsolve solution

Solving time : 0.001 (sec)
Leaf size : 8

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

\[ y = c_1 \,{\mathrm e}^{x} \]

1.8.7 Mathematica DSolve solution

Solving time : 0.023 (sec)
Leaf size : 16

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

\begin{align*} y(x)\to c_1 e^x \\ y(x)\to 0 \\ \end{align*}