The integrating factor of the differential equation \(y\displaystyle\frac{dx}{dy} + 3x = y^4\), \(y \gt 0\), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y^3\)
Here x is the dependent variable. Dividing by y gives the linear form \(\frac{dx}{dy} + P_1 x = Q_1\), where \(P_1\) and \(Q_1\) are functions of y only:
\( \displaystyle \frac{dx}{dy} + \frac{3}{y}\,x = y^3 \)
So
\( \displaystyle \int P_1\,dy = \int \frac{3}{y}\,dy = 3\log y \)
The integrating factor is
\( e^{3\log y} = y^3 \)
Why the other options are incorrect
Option (1) \(e^{3y}\) — Incorrect. This takes \(P_1 = 3\), forgetting to divide by y first.
Option (2) \(x^3\) — Incorrect. In \(\frac{dx}{dy} + P_1 x = Q_1\) the integrating factor is a function of y, never of x.
Option (3) \(\displaystyle\frac{1}{y^3}\) — Incorrect. This would need \(P_1 = -\frac{3}{y}\); here \(P_1\) is positive.