Question

Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]

Hint:

A first-order linear differential equation has the form \(\frac{dx}{dy}+P(y)x=Q(y)\). Solve it using the integrating factor \(e^{\int P(y)dy}\).

Answer 1

Step 1: Rewrite the equation.
\[ y\log y\frac{dx}{dy}+x=\frac{2}{y} \] Divide by \(y\log y\): \[ \frac{dx}{dy}+\frac{x}{y\log y}=\frac{2}{y^2\log y} \] This is a linear differential equation.
Step 2: Identify $P(y)$.
\[ P(y)=\frac{1}{y\log y} \] Thus the integrating factor (I.F.) is \[ \text{I.F.}=e^{\int P(y)\,dy} \] \[ =e^{\int \frac{1}{y\log y}\,dy} \] Let \[ t=\log y \] Then \[ dt=\frac{dy}{y} \] Thus \[ \int\frac{1}{y\log y}dy=\int\frac{1}{t}dt \] \[ =\log|\log y| \] Therefore \[ \text{I.F.}=e^{\log(\log y)}=\log y \] Step 3: Multiply the equation by I.F.
\[ (\log y)\frac{dx}{dy}+\frac{x}{y}=\frac{2}{y^2} \] The left side becomes a derivative: \[ \frac{d}{dy}(x\log y)=\frac{2}{y^2} \] Step 4: Integrate both sides.
\[ x\log y=\int\frac{2}{y^2}dy \] \[ x\log y=-\frac{2}{y}+C \] Step 5: Write the general solution.
\[ x\log y=-\frac{2}{y}+C \] Final Answer:
\[ \boxed{x\log y=-\frac{2}{y}+C} \]