伯努利微分方程1695

伯努利微分方程是形式如 y'+P(x)y=Q(x)yⁿ 的常微分方程。

解法： y'+P(x)y=Q(x)yⁿ

代入 w=y¹⁻ⁿ （注意 （公式略））： (w')/(1-n)+P(x)w=Q(x)

此一阶常微分方程可用积分因子求解。

例子

• 解以下微分方程。 y'-2y/x=-x²y²
• 两边除以 y²，得： y'y⁻²-2/xy⁻¹=-x²
• 利用分离变数法，可得： w=1/y
• （公式略）
• w'+2/xw=x²

它可以用积分因子的方法来解出。 M(x)=e^(2∫ 1/xdx)=x².

两边乘以 M(x)，得： w'x²+2xw=x⁴,

等式的左边是 wx² 的导数。两边积分，得： ∫ (wx²)'dx=∫ x⁴dx

• wx²=1/5x⁵+C
• 1/yx²=1/5x⁵+C
• 于是： （公式略）

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, whereas others require that n not be 0 or 1 as they cause the equation to become linear. 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.

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 n≠ 0 and n≠ 1, the substitution u=y¹⁻ⁿ reduces any Bernoulli equation to a linear differential equation, as shown: du/dx-(n-1)P(x)u=-(n-1)Q(x).

For example, in the case n=2, making the substitution u=y⁻¹ in the differential equation dy/dx+1/xy=xy² produces the equation du/dx-1/xu=-x, which is a linear differential equation.

Solution

Let x₀∈ (a,b) and （公式略）

be a solution of the linear differential equation z'(x)=(1-α)P(x)z(x)+(1-α)Q(x).

Then we have that y(x):=[z(x)]^(1/(1-α)) is a solution of y'(x)=P(x)y(x)+Q(x)y^(α)(x), y(x₀)=y₀:=[z(x₀)]^(1/(1-α)).

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

Example

Consider the Bernoulli equation y'-2y/x=-x²y²

(in this case, more specifically a Riccati equation).

The constant function y=0 is a solution.

Division by y² yields y'y⁻²-2/xy⁻¹=-x²

Changing variables gives the equations （公式略）

which can be solved using the integrating factor M(x)=e^(2∫ 1/x dx)=e^(2ln x)=x².

Multiplying by M(x), u'x²+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 （公式略）

The solution for y （公式略）