2.1.2 Problem 2

Internal problem ID [10260]
Book : First order enumerated odes
Section : section 1
Problem number : 2
Date solved : Thursday, November 27, 2025 at 10:28:27 AM
CAS classification : [_quadrature]

Solved using first_order_ode_quadrature

Time used: 0.049 (sec)

Solve

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

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

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

Summary of solutions found

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

Solved using first_order_ode_exact

Time used: 0.073 (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 (a\right )\mathop {\mathrm {d}x}\\ \left (-a\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) &= -a\\ 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 (-a\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 -a\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -a 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 \(\phi \)

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

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

\[ c_1 = -a x +y \]

Solving for \(y\) gives

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

Summary of solutions found

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

Solved using first_order_ode_homog_type_D2

Time used: 0.116 (sec)

Solve

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

Which is now solved The ode

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

is separable as it can be written as

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

Where

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

Integrating gives

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

\[ -\ln \left (-u \left (x \right )+a \right )=\ln \left (x \right )+c_1 \]

Taking the exponential of both sides the solution becomes

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

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

\[ -u +a=0 \]

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

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

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

\begin{align*} \frac {1}{-\frac {y}{x}+a} = c_1 x \end{align*}

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

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

Simplifying the above gives

\begin{align*} \frac {x}{a x -y} &= c_1 x \\ y &= a x \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= a x \\ y &= \frac {c_1 a x -1}{c_1} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= a x \\ y &= \frac {c_1 a x -1}{c_1} \\ \end{align*}

Solved using first_order_ode_homog_type_G

Time used: 0.059 (sec)

Solve

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

Multiplying the right side of the ode, which is \(a\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) a\\ &= \frac {x a}{y}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= \frac {x a}{y}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x a}{y}\\ \alpha &= \frac {f_x}{f_y} \\ &=-1 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G.

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x}} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=\frac {a}{z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1-\frac {a}{z}\right )}d z = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1-\frac {a}{z}\right )}d z &= 0 \\ \end{align*}

✓ Maple. Time used: 0.002 (sec). Leaf size: 9

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

\[ y = a x +c_1 \]

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

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

✓ Mathematica. Time used: 0.001 (sec). Leaf size: 11

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

\begin{align*} y(x)&\to a x+c_1 \end{align*}

✓ Sympy. Time used: 0.019 (sec). Leaf size: 7

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

\[ y{\left (x \right )} = C_{1} + a x \]