Differential Equations – Bernoulli’s Equation

Differential Equations – Bernoulli’s Equation

An equation of the form \(\frac{dy}{dx}+Py=Q{{y}^{n}}\). Where P and Q are Functions of X only and n ≠ 0, 1 is known as Bernoulli’s Equation.

It is easy to reduce the equation into linear form as below on dividing both sides by yⁿ. We get


Put y¹⁻ⁿ = z.

\(\left( 1-n \right){{y}^{1-n}}\frac{dy}{dx}=\frac{dz}{dx}\).

∴ Given Equation becomes \(\frac{dz}{dx}+\left( 1-n \right)Pz=\left( 1-n \right)Q\).

Which is a linear differential equation in z.

Here, \(IF={{e}^{\int{\left( 1-n \right)P.dx}}}\).

Required solution is \(z\left( IF \right)=\int{\left( 1-n \right).Q.{{e}^{\int{\left( 1-n \right)P.dx}}}dx}\).

Note: If equation is of the form \(\frac{dx}{dy}+Px=Q{{x}^{n}}\), where n ≠ 0, 1 and P and Q are functions of y only. Then divide by xⁿ and put x¹ ⁻ ⁿ = z.

\(\left( 1-n \right){{x}^{-n}}\frac{dx}{dy}=\frac{dz}{dx}\).

Example: What is the solution of \(x\left( \frac{dy}{dx} \right)+y={{y}^{2}}\log x\).

Solution: The given equation can be written as \(\frac{dy}{dx}+\frac{1}{x}y=\frac{{{y}^{2}}}{x}\log x\) (since dividing throughout by x).

(Or) \(\frac{1}{{{y}^{2}}}.\frac{dy}{dx}+\frac{1}{x}.\frac{1}{y}=\frac{\log x}{x}\) … (i) (since dividing throughout by y²)

Now, put 1/y = v, so that \(\left( -\frac{1}{{{y}^{2}}} \right)\frac{dy}{dx}=\left( \frac{dv}{dx} \right)\).

With these substitutions Equation (i), becomes

\(-\frac{dv}{dx}+\frac{1}{x}.v=\frac{\log x}{x}\).

Or \(\frac{dv}{dx}-\frac{1}{x}.v=-\frac{\log x}{x}\).

This is linear with v as the dependent variable.

Here, P = -1/x and Q = – (log x)/ x

∴ \(IF={{e}^{\int{Pdx}}}\),

\(={{e}^{\int{\left( -\frac{1}{x} \right)dx}}}\),

= e-log x

\(={{e}^{\log \left( \frac{1}{x} \right)}}\),

= 1/x.

Hence, the solution is \(v\left( \frac{1}{x} \right)=\int{\left\{ -\frac{\log x}{x} \right\}.\left( \frac{1}{x} \right)dx+C}\),

\(\frac{v}{x}=-\int{\left( \frac{1}{{{x}^{2}}} \right)\log x\,dx+C}\),

\(=-\left[ \left( -\frac{1}{x} \right)\log x-\int{\left( \frac{1}{x} \right)\left( -\frac{1}{x} \right)dx} \right]+C\) (integrating by parts 1/x² as the second function)

\(=\left( \frac{1}{x} \right)\log x-\int{\left( \frac{1}{{{x}^{2}}} \right)}dx+C\),

\(=\left( \frac{1}{x} \right)\log x+\frac{1}{x}+C\),

Or \(\frac{1}{\left( xy \right)}=\left( \frac{1}{x} \right)\left( 1+\log x \right)+C\),

Or 1 = y (1 + log x) + C xy.