2.1.46 Problem 46

Internal problem ID [10032]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 46
Date solved : Monday, December 08, 2025 at 07:08:54 PM
CAS classification : [_quadrature]

2.1.46.1 Solved using first_order_ode_autonomous

0.032 (sec)

Entering first order ode autonomous solver

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

Integrating gives

\begin{align*} \int f d f &= dx\\ \frac {f^{2}}{2}&= x +c_1 \end{align*}

Solving for \(f\) gives

\begin{align*} f &= \sqrt {2 x +2 c_1} \\ f &= -\sqrt {2 x +2 c_1} \\ \end{align*}

Figure 2.112: Slope field \(f^{\prime } = \frac {1}{f}\)

Summary of solutions found

\begin{align*} f &= \sqrt {2 x +2 c_1} \\ f &= -\sqrt {2 x +2 c_1} \\ \end{align*}

2.1.46.2 Solved using first_order_ode_bernoulli

0.990 (sec)

Entering first order ode bernoulli solver

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

In canonical form, the ODE is

\begin{align*} f' &= F(x,f)\\ &= \frac {1}{f} \end{align*}

This is a Bernoulli ODE.

\[ f' = \left (1\right ) \frac {1}{f} \tag {1} \]

The standard Bernoulli ODE has the form

\[ f' = f_0(x)f+f_1(x)f^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=0\\ f_1 &=1 \end{align*}

The first step is to divide the above equation by \(f^n \) which gives

\[ \frac {f'}{f^n} = f_1(x) \tag {3} \]

The next step is use the substitution \(v = f^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(f(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=0\\ f_1(x)&=1\\ n &=-1 \end{align*}

Dividing both sides of ODE (1) by \(f^n=\frac {1}{f}\) gives

\begin{align*} f'f &= 0 +1 \tag {4} \end{align*}

Let

\begin{align*} v &= f^{1-n} \\ &= f^{2} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= 2 ff' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} \frac {v^{\prime }\left (x \right )}{2}&= 1\\ v' &= 2 \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

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

\begin{align*} \int {dv} &= \int {2\, dx}\\ v \left (x \right ) &= 2 x + c_1 \end{align*}

The substitution \(v = f^{1-n}\) is now used to convert the above solution back to \(f\) which results in

\[ f^{2} = 2 x +c_1 \]

Solving for \(f\) gives

\begin{align*} f &= \sqrt {2 x +c_1} \\ f &= -\sqrt {2 x +c_1} \\ \end{align*}

Figure 2.113: Slope field \(f^{\prime } = \frac {1}{f}\)

Summary of solutions found

\begin{align*} f &= \sqrt {2 x +c_1} \\ f &= -\sqrt {2 x +c_1} \\ \end{align*}

2.1.46.3 Solved using first_order_ode_exact

0.099 (sec)

Entering first order ode exact solver

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

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

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

Therefore

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

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

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

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

Using result found above gives

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

And

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

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

\begin{align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial f } &= 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(f) \\ \end{align*}

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

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

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

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

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

\[ f'(f) = f \]

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

\begin{align*} \int f'(f) \mathop {\mathrm {d}f} &= \int \left ( f\right ) \mathop {\mathrm {d}f} \\ f(f) &= \frac {f^{2}}{2}+ c_1 \\ \end{align*}

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

\[ \phi = -x +\frac {f^{2}}{2}+ c_1 \]

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

\[ c_1 = -x +\frac {f^{2}}{2} \]

Solving for \(f\) gives

\begin{align*} f &= \sqrt {2 x +2 c_1} \\ f &= -\sqrt {2 x +2 c_1} \\ \end{align*}

Figure 2.114: Slope field \(f^{\prime } = \frac {1}{f}\)

Summary of solutions found

\begin{align*} f &= \sqrt {2 x +2 c_1} \\ f &= -\sqrt {2 x +2 c_1} \\ \end{align*}

2.1.46.4 Solved using first_order_ode_homog_type_G

0.077 (sec)

Entering first order ode homog type G solver

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

Multiplying the right side of the ode, which is \(\frac {1}{f}\) by \(\frac {x}{f}\) gives

\begin{align*} f^{\prime } &= \left (\frac {x}{f}\right ) \frac {1}{f}\\ &= \frac {x}{f^{2}}\\ &= F(x,f) \end{align*}

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

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , f\right )\right )\\ &= \frac {x}{f^{2}}\\ f_f&= f \left (\frac {\partial }{\partial f}F \left (x , f\right )\right )\\ &= -\frac {2 x}{f^{2}}\\ \alpha &= \frac {f_x}{f_f} \\ &=-{\frac {1}{2}} \end{align*}

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

Let

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

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

\begin{align*} F \left (z \right ) &=\frac {1}{z^{2}} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(f\). 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 ^{f x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {f}{\sqrt {x}}}\frac {1}{z \left (\frac {1}{2}-\frac {1}{z^{2}}\right )}d z = 0 \]

Figure 2.115: Slope field \(f^{\prime } = \frac {1}{f}\)

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 +\int _{}^{\frac {f}{\sqrt {x}}}\frac {1}{z \left (\frac {1}{2}-\frac {1}{z^{2}}\right )}d z &= 0 \\ \end{align*}

2.1.46.5 ✓ Maple. Time used: 0.002 (sec). Leaf size: 23

ode:=diff(f(x),x) = 1/f(x); 
dsolve(ode,f(x), singsol=all);
 

\begin{align*} f &= \sqrt {2 x +c_1} \\ f &= -\sqrt {2 x +c_1} \\ \end{align*}

Maple trace

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

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}f \left (x \right )=\frac {1}{f \left (x \right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}f \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}f \left (x \right )=\frac {1}{f \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}f \left (x \right )\right ) f \left (x \right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}f \left (x \right )\right ) f \left (x \right )d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {f \left (x \right )^{2}}{2}=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} f \left (x \right ) \\ {} & {} & \left \{f \left (x \right )=\sqrt {2 \mathit {C1} +2 x}, f \left (x \right )=-\sqrt {2 \mathit {C1} +2 x}\right \} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{f \left (x \right )=\sqrt {\mathit {C1} +2 x}, f \left (x \right )=-\sqrt {\mathit {C1} +2 x}\right \} \end {array} \]

2.1.46.6 ✓ Mathematica. Time used: 0.036 (sec). Leaf size: 38

ode=D[ f[x],x]==f[x]^(-1); 
ic={}; 
DSolve[{ode,ic},f[x],x,IncludeSingularSolutions->True]
 

\begin{align*} f(x)&\to -\sqrt {2} \sqrt {x+c_1}\\ f(x)&\to \sqrt {2} \sqrt {x+c_1} \end{align*}

2.1.46.7 ✓ Sympy. Time used: 0.131 (sec). Leaf size: 22

from sympy import * 
x = symbols("x") 
f = Function("f") 
ode = Eq(Derivative(f(x), x) - 1/f(x),0) 
ics = {} 
dsolve(ode,func=f(x),ics=ics)
                                                                                  
                                                                                  
 

\[ \left [ f{\left (x \right )} = - \sqrt {C_{1} + 2 x}, \ f{\left (x \right )} = \sqrt {C_{1} + 2 x}\right ] \]