# 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

$${{y}^{-n}}\frac{dy}{dx}+P{{y}^{1-n}}=Q$$.

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.