Consider the following differential equations:
(A) \(\displaystyle\frac{dy}{dx} + x^2y = \sin x\)
(B) \(\displaystyle\frac{dy}{dx} + y^2 = x\)
(C) \(\displaystyle\frac{dx}{dy} + x = y^2\)
(D) \(\displaystyle\frac{dy}{dx} = \frac{1}{e^{-y} - x}\)
Which of the above are linear differential equations, either of the form \(\displaystyle\frac{dy}{dx} + Py = Q\) (P, Q functions of x) or of the form \(\displaystyle\frac{dx}{dy} + P_1x = Q_1\) (\(P_1\), \(Q_1\) functions of y)?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
A linear equation contains the dependent variable and its derivative only to the first power, and the coefficients depend on the other variable only.
In (D), the reciprocal of both sides gives
\( \displaystyle \frac{dx}{dy} = e^{-y} - x \)
\( \displaystyle \frac{dx}{dy} + x = e^{-y} \)
So (A) is linear in y, (C) and (D) are linear in x, and (B) contains \(y^2\).
Checking the statements
(A) \(\frac{dy}{dx} + x^2y = \sin x\) — Correct. Linear in y with \(P = x^2\), \(Q = \sin x\).
(B) \(\frac{dy}{dx} + y^2 = x\) — Incorrect. The term \(y^2\) is not of the first power in y.
(C) \(\frac{dx}{dy} + x = y^2\) — Correct. Linear in x with \(P_1 = 1\), \(Q_1 = y^2\).
(D) \(\frac{dy}{dx} = \frac{1}{e^{-y} - x}\) — Correct. It becomes \(\frac{dx}{dy} + x = e^{-y}\), linear in x.