Consider the differential equation \(\displaystyle\frac{dx}{dy} + x = 2y\).
(A) It is a homogeneous differential equation.
(B) Its integrating factor is \(e^y\).
(C) Its general solution is \(x = 2y - 2 + Ce^{-y}\), where C is an arbitrary constant.
(D) For the solution with \(x = 0\) when \(y = 0\), the value of x at \(y = 1\) is \(\displaystyle\frac{2}{e}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (C) and (D) only
Here x is the dependent variable. The equation is linear in x, \(\frac{dx}{dy} + P_1x = Q_1\), with \(P_1 = 1\) and \(Q_1 = 2y\).
The integrating factor is
\( e^{\int 1\,dy} = e^y \)
So
\( \displaystyle x\,e^y = \int 2y\,e^y\,dy \)
Integrating by parts:
\( x\,e^y = 2(y - 1)e^y + C \)
\( x = 2y - 2 + Ce^{-y} \)
With \(x = 0\) at \(y = 0\): \(0 = -2 + C\), so \(C = 2\). At \(y = 1\):
\( \displaystyle x = 2 - 2 + 2e^{-1} = \frac{2}{e} \)
The equation is not homogeneous: \(\frac{dx}{dy} = 2y - x\), and \(2y - x\) has degree 1, not degree zero.
So the equation is not homogeneous, its integrating factor is \(e^y\), its general solution is \(x = 2y - 2 + Ce^{-y}\), and the particular solution gives \(x = \frac{2}{e}\) at \(y = 1\).
Checking the statements
(A) It is a homogeneous differential equation. — Incorrect. \(2y - x\) is of degree 1, not degree zero.
(B) Its integrating factor is \(e^y\). — Correct. \(P_1 = 1\), so the integrating factor is \(e^{\int 1\,dy} = e^y\).
(C) Its general solution is \(x = 2y - 2 + Ce^{-y}\). — Correct. \(x\,e^y = 2(y - 1)e^y + C\).
(D) For the solution with \(x = 0\) when \(y = 0\), the value of x at \(y = 1\) is \(\frac{2}{e}\). — Correct. \(C = 2\), so \(x(1) = 2e^{-1}\).