Question

Solution of the differential equation 6y^2 dx – x(x^3 + 2y)dy = 0 is ________

(a) \(\frac{y}{x^3} = \frac{-log⁡y}{2} + c \)

(b) \(\frac{y^2}{x^3} = \frac{-log⁡x}{4} + c  \)

(c) \(\frac{x}{y^3} = \frac{-log⁡x}{2} + c \)

(d) \(\frac{x}{y^2} = \frac{-log⁡y}{4} + c \)

I got this question by my school teacher while I was bunking the class.

My question is taken from Bernoulli Equations in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (a) \(\frac{y}{x^3} = \frac{-log⁡y}{2} + c \)

Explanation: Equation is reduced to \(\frac{dy}{dx} = \frac{x(x^3+2y)}{6y^2} \,i.e\, \frac{dx}{dy}-\frac{x}{3y} = \frac{x^4}{6y^2}\) …divide by x^4

we get \(\frac{1}{x^4} \frac{dy}{dx} – \frac{1}{3x^3 y} = \frac{1}{6y^2} \,put\, \frac{1}{x^3} = t \rightarrow \frac{-3}{x^4} \frac{dx}{dy} = \frac{dt}{dy} \,or\, \frac{1}{x^4} \frac{dy}{dx} = \frac{-1}{3} \frac{dt}{dy} \)

substituting \(\frac{-1}{3} \frac{dt}{dy}  – \frac{t}{3y} = \frac{1}{6y^2}  \,or\, \frac{dt}{dy} + \frac{t}{y} = \frac{-1}{2y^2}\) this equation is linear in t i.e it is of the form

\(\frac{dt}{dy}  + Pt = Q \,where\, P=\frac{1}{y}, Q=\frac{-1}{2y^2} \,I.F\, = e^{\int P \,dy} = e^{\int \frac{1}{y} \,dy} = e^{log⁡y} = y\) its solution is \(te^{\int P \,dy} = \int Q \,e^{\int P \,dy} + c \rightarrow ty = \int y * \frac{-1}{2y^2} \,dy + c = \frac{-log⁡y}{2} + c\)

\(\frac{y}{x^3} = \frac{-log⁡y}{2} + c \) is the required solution.