2.1.10 Problem 10

Internal problem ID [10268]
Book : First order enumerated odes
Section : section 1
Problem number : 10
Date solved : Thursday, November 27, 2025 at 10:28:41 AM
CAS classification : [_quadrature]

Solved using first_order_ode_autonomous

Time used: 0.026 (sec)

Solve

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

Integrating gives

\begin{align*} \int \frac {1}{y b}d y &= dx\\ \frac {\ln \left (y \right )}{b}&= x +c_1 \end{align*}

Singular solutions are found by solving

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

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

The following diagram is the phase line diagram. It classifies each of the above equilibrium points as stable or not stable or semi-stable.

Figure 2.15: Phase line diagram

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{c_1 b +x b} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{c_1 b +x b} \\ \end{align*}

Solved using first_order_ode_exact

Time used: 0.113 (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 b\right )\mathop {\mathrm {d}x}\\ \left (-y b\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 b\\ 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 b\right )\\ &= -b \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 ( -b\right ) - \left (0 \right ) \right ) \\ &=-b \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 -b\mathop {\mathrm {d}x} } \end{align*}

The result of integrating gives

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

And

\begin{align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{-x b}\left (1\right ) \\ &= {\mathrm e}^{-x b} \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 b \,{\mathrm e}^{-x b}\right ) + \left ({\mathrm e}^{-x b}\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 b}\mathop {\mathrm {d}y} \\ \tag{3} \phi &= y \,{\mathrm e}^{-x b}+ 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 b \,{\mathrm e}^{-x b}+f'(x) \end{equation}

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

\begin{equation} \tag{5} -y b \,{\mathrm e}^{-x b} = -y b \,{\mathrm e}^{-x b}+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 \(\phi \)

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

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

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

Solving for \(y\) gives

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

Summary of solutions found

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

Solved using first_order_ode_dAlembert

Time used: 0.064 (sec)

Solve

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

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} p = y b \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= \frac {p}{b} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= \frac {p}{b} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \frac {p^{\prime }\left (x \right )}{b} \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

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

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = p \left (x \right ) b \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {1}{p b}d p &= dx\\ \frac {\ln \left (p \right )}{b}&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} p b&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= \frac {{\mathrm e}^{c_1 b +x b}}{b} \\ y &= 0 \\ \end{align*}

Simplifying the above gives

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

Summary of solutions found

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

Solved using first_order_ode_homog_type_D2

Time used: 0.080 (sec)

Solve

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

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 b \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (x b -1\right )}{x} \end{equation}

is separable as it can be written as

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

Where

\begin{align*} f(x) &= \frac {x b -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 b -1}{x} \,dx} \\ \end{align*}

\[ \ln \left (u \left (x \right )\right )=x b +\ln \left (\frac {1}{x}\right )+c_1 \]

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

\[ u=0 \]

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

Converting \(\ln \left (u \left (x \right )\right ) = x b +\ln \left (\frac {1}{x}\right )+c_1\) back to \(y\) gives

\begin{align*} \ln \left (\frac {y}{x}\right ) = x b +\ln \left (\frac {1}{x}\right )+c_1 \end{align*}

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

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

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{x b +c_1} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{x b +c_1} \\ \end{align*}

✓ Maple. Time used: 0.002 (sec). Leaf size: 10

ode:=diff(y(x),x) = b*y(x); 
dsolve(ode,y(x), singsol=all);
 

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

Maple trace

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

Maple step by step

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

✓ Mathematica. Time used: 0.015 (sec). Leaf size: 18

ode=D[y[x],x]==b*y[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

✓ Sympy. Time used: 0.058 (sec). Leaf size: 8

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

\[ y{\left (x \right )} = C_{1} e^{b x} \]