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
\(=\int{t\times {{e}^{t}}}.\frac{dt}{2}\),
\(=\frac{1}{2}t\int{{{e}^{t}}}.dt-\frac{1}{2}\int{{{e}^{t}}}.dt\),
\(=\frac{1}{2}t{{e}^{t}}-\frac{1}{2}{{e}^{t}}+C\),
\(=\frac{1}{2}{{(x)}^{2}}{{e}^{{{x}^{2}}}}-\frac{1}{2}{{e}^{{{x}^{2}}}}+C\).