Bernoulli Equations

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

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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

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Example 4: (Logistic equation)

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To plot the results, we employ two options:

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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,

k := 3: 
a := 1:
y := x-> 2*exp(x/(2*a))*sqrt(a*(a+x)*exp(-x/a)- a^2 + k^2/4):
the_parabola := a*x:
xA := 1:
yA := y(xA):
yPrime:= x-> diff(y(x), x):
kSlope := -1/eval(yPrime(x),x=xA):
bIntercept := yA - kSlope*xA:
xB := -bIntercept/kSlope:
yB := 0:
xC := (xA + xB)/2:
yC := yA/2:

while abs(yC^2 - eval(the_parabola,x=xC)) > 1/1000 do
   xC := xC + 1/100; 
   yC := sqrt(eval(the_parabola,x=xC));
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p1:=plottools:-line([xA,yA],[xB,yB],color="Blue"):
p2:=plot([y(x), sqrt(the_parabola)],x=-5..11, y=0..6):
plots:-display([p1,p2]);

the_parabola := a*x:

Choose a point A on the curve y = y(x)

xA := 1:
yA := y(xA):

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kSlope;
evalf(%);

kSlope;
evalf(%);

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is( Student[Precalculus]:-Distance([xA, yA], [xC, yC]) = Student[Precalculus]:-Distance([xB, yB], [xC, yC]));

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Is C at the midpoint of the line?

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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).

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

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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:

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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.

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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?