Bernoulli differential equation

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Short description: Type of ordinary differential equation

In mathematics, an ordinary differential equation is called a Bernoulli differential equation if it is of the form

y+P(x)y=Q(x)yn,

where n is a real number. Some authors allow any real n,[1][2] whereas others require that n not be 0 or 1.[3][4] The equation was first discussed in a work of 1695 by Jacob Bernoulli, after whom it is named. The earliest solution, however, was offered by Gottfried Leibniz, who published his result in the same year and whose method is the one still used today.[5]

Bernoulli equations are special because they are nonlinear differential equations with known exact solutions. A notable special case of the Bernoulli equation is the logistic differential equation.

Transformation to a linear differential equation

When n=0, the differential equation is linear. When n=1, it is separable. In these cases, standard techniques for solving equations of those forms can be applied. For n0 and n1, the substitution u=y1n reduces any Bernoulli equation to a linear differential equation

dudx(n1)P(x)u=(n1)Q(x).

For example, in the case n=2, making the substitution u=y1 in the differential equation dydx+1xy=xy2 produces the equation dudx1xu=x, which is a linear differential equation.

Solution

Let x0(a,b) and

{z:(a,b)(0,) ,if α{1,2},z:(a,b){0} ,if α=2,

be a solution of the linear differential equation

z(x)=(1α)P(x)z(x)+(1α)Q(x).

Then we have that y(x):=[z(x)]11α is a solution of

y(x)=P(x)y(x)+Q(x)yα(x) , y(x0)=y0:=[z(x0)]11α.

And for every such differential equation, for all α>0 we have y0 as solution for y0=0.

Example

Consider the Bernoulli equation

y2yx=x2y2

(in this case, more specifically a Riccati equation). The constant function y=0 is a solution. Division by y2 yields

yy22xy1=x2

Changing variables gives the equations

u=1y,u=yy2u2xu=x2u+2xu=x2

which can be solved using the integrating factor

M(x)=e21xdx=e2lnx=x2.

Multiplying by M(x),

ux2+2xu=x4.

The left side can be represented as the derivative of ux2 by reversing the product rule. Applying the chain rule and integrating both sides with respect to x results in the equations

(ux2)dx=x4dxux2=15x5+C1yx2=15x5+C

The solution for y is

y=x215x5+C.

Notes

References

  • Bernoulli, Jacob (1695), "Explicationes, Annotationes & Additiones ad ea, quae in Actis sup. de Curva Elastica, Isochrona Paracentrica, & Velaria, hinc inde memorata, & paratim controversa legundur; ubi de Linea mediarum directionum, alliisque novis", Acta Eruditorum . Cited in (Hairer Nørsett).
  • Hairer, Ernst; Nørsett, Syvert Paul; Wanner, Gerhard (1993), Solving ordinary differential equations I: Nonstiff problems, Berlin, New York: Springer-Verlag, ISBN 978-3-540-56670-0 .