Consider the following statements:
(A) The integrating factor of \(\displaystyle\frac{dy}{dx} + Py = Q\), where P and Q are functions of x, is \(e^{\int P\,dx}\).
(B) The general solution of \(\displaystyle\frac{dx}{dy} + P_1x = Q_1\), where \(P_1\) and \(Q_1\) are functions of y, is \(x \cdot e^{\int P_1\,dy} = \int \left(Q_1 \cdot e^{\int P_1\,dy}\right) dy + C\).
(C) The differential equation \(\displaystyle\frac{dy}{dx} = F(x, y)\) is homogeneous when F(x, y) is a homogeneous function of degree one.
(D) A homogeneous differential equation of the form \(\displaystyle\frac{dy}{dx} = g\left(\frac{y}{x}\right)\) can be solved by the substitution \(y = vx\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
For \(\frac{dy}{dx} + Py = Q\), multiplying by \(e^{\int P\,dx}\) turns the left side into the derivative of \(y \cdot e^{\int P\,dx}\), which gives
\( \displaystyle y \cdot e^{\int P\,dx} = \int \left(Q \cdot e^{\int P\,dx}\right) dx + C \)
The form \(\frac{dx}{dy} + P_1x = Q_1\) is the same with x and y interchanged, so its integrating factor is \(e^{\int P_1\,dy}\).
The equation \(\frac{dy}{dx} = F(x, y)\) is homogeneous when F is a homogeneous function of degree zero; then \(F = g\left(\frac{y}{x}\right)\), and \(y = vx\) separates the variables.
So both linear forms use e raised to the integral of the coefficient, the degree needed for a homogeneous equation is zero, and y = vx is the right substitution.
Checking the statements
(A) The integrating factor of \(\frac{dy}{dx} + Py = Q\) is \(e^{\int P\,dx}\). — Correct. This is the standard integrating factor.
(B) The general solution of \(\frac{dx}{dy} + P_1x = Q_1\) is \(x \cdot e^{\int P_1\,dy} = \int \left(Q_1 \cdot e^{\int P_1\,dy}\right) dy + C\). — Correct. Same rule with x and y interchanged.
(C) The equation \(\frac{dy}{dx} = F(x, y)\) is homogeneous when F is homogeneous of degree one. — Incorrect. F must be homogeneous of degree zero.
(D) \(\frac{dy}{dx} = g\left(\frac{y}{x}\right)\) can be solved by the substitution \(y = vx\). — Correct. It gives \(x\frac{dv}{dx} = g(v) - v\), which is separable.