Question

Solution of the differential equation \((x+3y^2)\frac{dy}{dx} = y(y>0)\) is ______________

(a) x=3y^2+cy

(b) y=2x^2+c

(c) x=2y^2+\(\frac{c}{y}\)

(d) y=3x^3+c

I have been asked this question in unit test.

I would like to ask this question from First Order Linear Differential Equations topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) x=3y^2+cy

The best I can explain: \((x+3y^2)\frac{dy}{dx} = y\) can be rearranged to \(\frac{dx}{dy} = \frac{x}{y} + 3y \rightarrow \frac{dx}{dy} – \frac{x}{y} = 3y \)

above equation is of the form \(\frac{dx}{dy}\) + Px = Q where P & Q is a function of y only

given DE is linear DE in x here\( P = \frac{-1}{y}, Q=3y, e^{\int P \,dy} = e^{\int \frac{-1}{y} \,dy} = e^{log⁡y^{-1}}  = \frac{1}{y}\)

therefore its solution is given by \(xe^{\int P \,dy} = \int Q \,e^{\int P \,dy} dy + c\)

\(x \frac{1}{y} = \int 3y * \frac{1}{y}  \,dy + c = 3y + c\)

i.e x=3y^2+cy.