Example 1: Consider the differential equation \( \quad y\, y' = y^2 + e^x . \ \)   Since in the given differential equation the derivative is not isolated, we divide both sides by y. As a result, we obtain a Bernoulli equation y′ = y + e^x/y (here prime indicates differentiation: y′ = dy/dx). To solve it, we first use the Leibniz substitution:   u = y².   \( \displaystyle \quad \Longleftrightarrow \quad y = u^{1/2} . \quad \) Then \[ y' = \frac{1}{2}\, u^{-1/2} u' \quad \Longrightarrow \quad y\,y' = \frac{1}{2}\, u' \] and we get the linear differential equation with respect to unknown function u:

To solve it, we first find an integrating factor by solving the corresponding separable equation Multiplying both sides of the equation \( u' - 2\,u = 2\, e^x \) by the integrating factor, we obtain an exact equation where C is a constant of integration. Returning to the original function, we find the general solution to the given differential equation \( y = \pm \sqrt{C\,e^{2x} - 2\, e^x} . \)

Now we demonstrate the application of the Bernoulli method by seeking the solution as the product y = u v, where u is a solution of the "linear part:"

Then v must be the general solution to the separable equation Separating variables and integrating, we obtain Since \( v = \pm\sqrt{C- 2\,e^{-x}} , \) we multiply it by u to obtain the general solution. Now we use Mathematica:

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Example 2: The four streamlines (corresponding to the values of an arbitrary constant C=1,2,3,4) from the general solution y=1/(x Sqrt[C-2 ln[x]]) of the Bernoulli equation
x y' =x^2 y^3 -y can be plotted with one command:

and the corresponding direction field:

Some trajectories.		Direction field.

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Example 3:: Consider the Bernoulli equation \( y' +x\,y=x\, y^4 . \) Using Leibniz substitution

we reduce the given differential equation to a linear one.

To find an integrating factor for the latter, we need to solve the separable equation Multiplication of the differential equation \( u' - 3x \, u = -3x \) by μ yields an exact equation Since \( y = u^{-1/3} , \) we obtain the general solution: \[ y(x) = \left( 1 + C\, e^{3 x^2 /2} \right)^{-1/3} . \]

We employ Mathematica to confirm our analytical manipulations. First, we define type in the Bernoulli equation and male Leibniz's substitution. < We substitute into the differential equation Find the integrating factor Solve the linear differential equation Back substitution to find y

Now we find the general solution using the Bernoulli method: y = u·v, where u is a solution of the "truncated" linear equation \( u' + x\,u =0 \) and v is the general solution of the separable equation \( u\, v' = x\, \left( u\, v \right)^4 . \) Solving the equation for u, we get

Using this result, we return to the equation for v: Multiplying u and v, we obtain the general solution of the given differential equation: Mathematica confirms our solution: This answer is slightly different from the general solution obtained by Leibniz's method: it contains a multiple of "3" as a multiple of an arbitrary constant. Clearly, we can denote 3 c as a new constant, so both methods provide the same solutiuon.    ■

Example 4: (Logistic equation)

The logistic equation with harvesting \( \dot{y} = r\, y \left( 1 - \frac{y}{k} \right) - h\, y(t) \) is an example of the Bernoulli equation. It is assumed that all coefficients (r, k, h) are constants. It can be solved and plotted as follows. We define the solution function

To plot the results, we employ two options:

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Example 5: Let us consider the curve y = y(x) that goes through point (0, K), K ≠ 0, on the ordinate. Suppose that the middle point of the normal to an arbitrary point on the curve and horizontal axis (abscissa) belongs to parabola y² = 𝑎x. Find the equation of the curve.

Solution: The slope of the tangent line to the curve is its derivative, y′(x) at any point x. First, we plot the picture. Setting some parameters,

we define the function y(x): Define the parabola y² = 𝑎 x Choose a point A on the curve y = y(x) Calculate the derivative of y(x) Calculate the slope of the normal line Calculate the intercept of the normal line Calculate point B on the x-axis Calculate midpoint C of AB Ensure midpoint C satisfies the parabola equation Plot the picture:

Is C at the midpoint of the line?

To determine whether Line ACB is the normal line to the curve y = y(x) at point A, the slope of Line ACB is the negative reciprocal of the slope of the tangent line to the curve at point A, which is kSlope. Calculate the intercept of the normal line Define the normal line equation. Slope is the same as kSlope

The slope of the normal line is k = −1/y′(x). With this in hand, we find the equation of the normal line y = k x + b that goes through an arbitrary point A(x, y) on the curve and crosses abscissa at B(X, 0). Since X belongs to the normal line, we obtain \[ 0 = k\,X + b \qquad \Longrightarrow \qquad X = - \frac{b}{k} = b\,y' (x) . \] Substituting y − k x instead of b, yields \[ X = - \frac{y-k\,x}{k} = x - \frac{y}{k} = x + y \cdot y' \] because k = −1/y′. The middle point C of the segment A B has coordinates (X₁, Y₁) that are mean values of corresponding coordinates A(x, y) and B(x + y·y′, 0), that is, \[ X_1 = \frac{x + x + y\,y'}{2} = x + \frac{y}{2} \, y' \quad\mbox{and} \quad Y_1 = \frac{y}{2} . \] The coordinates of point C satisfy the equation of the parabola, y² = 𝑎x. Thus, \[ Y_1^2 = \left( \frac{y}{2} \right)^2 = a\, X_1 \qquad \Longrightarrow \qquad X_1 = x + \frac{y}{2} \, y' = \frac{1}{a} \left( \frac{y}{2} \right)^2 \] or \[ \frac{{\text d} y}{{\text d}x} = \frac{y}{2a} - \frac{2x}{y} \tag{5.1} \] This is a Bernoulli equation. We plot the corresponding direction field and some typical solutions to Eq.(5.1).

Using the Bernoulli substitution y = u v, we obtain two separable differential equations, one for u \[ \frac{{\text d} u}{{\text d}x} = \frac{u}{2a} \qquad \Longrightarrow \qquad u(x) = e^{x/(2a)} , \tag{5.2} \] and another one for v, \[ u\,\frac{{\text d} v}{{\text d}x} = - \frac{2x}{u\,v} \qquad \Longrightarrow \qquad v\,\frac{{\text d} v}{{\text d}x} = - 2x\, e^{-x/a} . \tag{5.3} \] Solving Eq.(5.2), we set a constant of integration to 1 because it does not matter; it is important that this constant is not zero. Integrating both sides of Eq.(5.3), we obtain \[ \int v\,{\text d}v = \frac{v^2}{2} = - \int 2x\,e^{-x/a} \,{\text d}x = 2a \left( a+x \right) e^{-x/a} + c . \] We choose a constant of integration c so that v(0) = K, so \[ \frac{1}{2}\,v(0)^2 = 2a^2 + c = K^2 /2 > 0 \qquad \Longrightarrow \qquad c = \frac{K^2}{2} -2a^2 . \] This leads to \[ v^2 (x) = 4a \left( a+x \right) e^{-x/a} - 4a^2 + K^2 = 4a \left[ \left( a+x \right) e^{-x/a} - a \right] + K^2 . \] Taking square root, we obtain two possible formulas for function v: \[ v(x) = \pm 2\, \sqrt{a \left( a+x \right) e^{-x/a} - a^2 + K^2 /4} \tag{5.4} \] We plot one positive branch of function v(x) for K = 3.

The following animation shows the function v(x) depending on the value of K.

Multiplying expression (5.4) by u(x), we get the explicit formulas for two branches of the solution: \[ y(x) = \pm 2\, e^{x/(2a)} \, \sqrt{a \left( a+x \right) e^{-x/a} - a^2 + K^2 /4} = \pm 2\, \sqrt{a \left( a+x \right) - a^2 e^{x/a} + e^{x/a} \,K^2 /4} . \tag{5.4} \] We check with Mathematica:

Now capabilities of a computer algebra systems (such as Mathematica) are so powerful that we can check equivalence of many transcendent expressions with one line command. Previously, people verified equivalence of two expressions (one is obtained using standard command DSolve and another one was derived manually based on the Bernoulli method) were forced to evaluate these expressions at some numerical values of parameters. For example, if we were 20 years ago, we would check these two answers numerically upon evaluating the corresponding expressions at some points; for instance, we evaluate at x = 1.55.

and Since these two expressions have the same numerical value, we conclude that our solution is correct and Mathematica is not ready yet to do the corresponding arithmetic manipulations.

We plot both branches of the solution for two particular values of parameters: 𝑎 = 1 and K = 3.

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1. Find the general solution of the logistic equation:    y′ = y(2 - 5y).
2. Find a particular solution to the initial value problem:    y′ = y²e^x − y,    y(0) = 1.
3. Find the general solution of the Bernoulli equation:    y′ = xy³ − 3y.
4. Find a particular solution to the initial value problem:    y′ = y²e^3x + 4y,    y(0) = −1.
5. Find the general solution of the Bernoulli equation:    y′ + 2 y + 4 xy³ = 0.
6. For the Bernoulli equation    y′ = y³e^x − 2y, find the positive solution that separates solutions tending to zero from those having a vertical asymptote at a positive value of x. What is the initial value y(0) for this solution?

1. Azevedo, D. and Valentino, M.C., Generalization of the Bernoulli ODE, International Journal of Mathematical Education in Science and Technology, 2017, Volume 48, Issue 2, pp. 256--260; doi:https://doi.org/10.1080/0020739X.2016.1201599
2. Tisdell, C.C., Alternate solution to generalized Bernoulli equations via an integrating factor: an exact differential equation approach, International Journal of Mathematical Education in Science and Technology, 2017, Volume 48, Issue 6, pp. 913--918; doi: https://doi.org/10.1080/0020739X.2016.1272143