# Bernoulli’s Equation

## Bernoulli’s Equation

Solution of Bernoulli’s Equation: Given equation is dy/ dx + Py = Qyⁿ

1/ yⁿ dy/ dx + P. 1/ yⁿ¯¹ = Q.

$$\frac{1}{{{y}^{n-1}}}=v$$, then $$\frac{-(n-1)}{{{y}^{n}}}\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow \frac{1}{{{y}^{n}}}\frac{dy}{dx}=\frac{1}{1-n}\frac{dv}{dx}$$,

Becomes $$\frac{1}{1-n}\frac{dv}{dx}+Pv=Q$$,

$$\frac{dv}{dx}+(1-n)Pv=(1-n)Q$$,

It is linear in v and can be solved.

Example: Solve $$\frac{dy}{dx}+x\sin 2y={{x}^{3}}{{\cos }^{2}}y$$.

Solution: Given that $$\frac{dy}{dx}+x\sin 2y={{x}^{3}}{{\cos }^{2}}y$$,

$$\frac{1}{{{\cos }^{2}}y}\times \frac{dy}{dx}+x\sin 2y\times \frac{1}{{{\cos }^{2}}y}={{x}^{3}}$$,

$${{\sec }^{2}}y\times \frac{dy}{dx}+x\sin 2y\times {{\sec }^{2}}y={{x}^{3}}$$,

$${{\sec }^{2}}y\times \frac{dy}{dx}+x\times 2\sin y\cos y\times {{\sec }^{2}}y={{x}^{3}}$$,

$${{\sec }^{2}}y\times \frac{dy}{dx}+x\tan y={{x}^{3}}$$ … (1)

Put tany = v

Differentiation with respect to ‘x’

sec²y dy/dx = dv/dx

dv/dx + 2xv = x³

it is linear in v. here P = 2x, Q = x³

Integration factor (I.F) $$={{e}^{\int{2xdx}}}$$.

$$={{e}^{2\times \frac{{{x}^{2}}}{2}}}={{e}^{{{x}^{2}}}}$$.

y x Integration factor (I.F) = $$\int{{{x}^{3}}\times {{e}^{{{x}^{2}}}}}dx$$.

Put x² = t

Differentiation with respect to ‘x’

2x = dt/dx

2x dx = dt

x. dx = dt/ 2

$$=\int{t\times {{e}^{t}}}.\frac{dt}{2}$$,

$$=\frac{1}{2}t\int{{{e}^{t}}}.dt-\frac{1}{2}\int{{{e}^{t}}}.dt$$,

$$=\frac{1}{2}t{{e}^{t}}-\frac{1}{2}{{e}^{t}}+C$$,

$$=\frac{1}{2}{{(x)}^{2}}{{e}^{{{x}^{2}}}}-\frac{1}{2}{{e}^{{{x}^{2}}}}+C$$.