Question

If \(y=y(x)\) is a solution of the differential equation, \(x\frac{dy}{dx}=y(2-3xy),\ x>0\) with \(y(1)=1\), then \(y\left(\frac{1}{4}\right)\) is equal to ________.

Hint:

Divide by \(y^2\) and put \(u=\frac1y\); the equation becomes linear in \(u\): \(u'+\frac{2}{x}u=3\).

Answer 1

Correct Answer: 4

Given \(x\frac{dy}{dx}=2y-3xy^2\).
Divide by \(y^2\): \(\frac{x}{y^2}\frac{dy}{dx}=\frac{2}{y}-3x\).
Let \(u=\frac1y\), so \(\frac{du}{dx}=-\frac{1}{y^2}\frac{dy}{dx}\). Then \(-x\frac{du}{dx}=2u-3x\), i.e. \(\frac{du}{dx}+\frac{2}{x}u=3\).
Integrating factor \(=e^{\int 2/x\,dx}=x^2\), so \(\frac{d}{dx}(x^2u)=3x^2\) and \(x^2u=x^3+C\).
Since \(y(1)=1\), \(u(1)=1\): \(1=1+C\), so \(C=0\).
Hence \(u=x\), i.e. \(y=\frac1x\).
Therefore \(y\left(\frac14\right)=4\).
Answer: 4