For a given differential equation \( e^{dy/dx}=x+1;x\in(-1,\infty) \), then which of the following statements are TRUE?
(A) Its general solution is \( y=x\log_e(x+1)-x+\log_e(x+1)+\mathrm{C} \): C is an arbitrary constant.
(B) Its particular solution is \( y=(x+1)\log_e(x+1)-x+5:y(0)=5 \)
(C) Its general solution is \( y=(x-1)\log(x+1)-x+\mathrm{C} \): C is an arbitrary constant
(D) Its particular solution is \( y=(x-1)\log(x+1)-x+5;y(0)=5 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (B) only
Taking logarithms of both sides,
\( \displaystyle \frac{dy}{dx}=\log_e(x+1) \)
Integrating by parts, with \( \log_e(x+1) \) as the first function and 1 as the second,
\( \displaystyle y=x\log_e(x+1)-\int\frac{x}{x+1}\,dx \)
\( \displaystyle =x\log_e(x+1)-\int\left(1-\frac{1}{x+1}\right)dx \)
\( \displaystyle =x\log_e(x+1)-x+\log_e(x+1)+\mathrm{C} \)
\( \displaystyle =(x+1)\log_e(x+1)-x+\mathrm{C} \)
Putting y(0) = 5:
\( \displaystyle 1\cdot\log_e 1-0+\mathrm{C}=5 \)
\( \displaystyle \mathrm{C}=5 \)
So the general solution is \( y=(x+1)\log_e(x+1)-x+\mathrm{C} \) and the particular solution with y(0) = 5 is \( y=(x+1)\log_e(x+1)-x+5 \).
Checking the statements
(A) Its general solution is \( y=x\log_e(x+1)-x+\log_e(x+1)+\mathrm{C} \): C is an arbitrary constant. — Correct. \( x\log_e(x+1)+\log_e(x+1)=(x+1)\log_e(x+1) \).
(B) Its particular solution is \( y=(x+1)\log_e(x+1)-x+5:y(0)=5 \) — Correct. This is the general solution with C = 5.
(C) Its general solution is \( y=(x-1)\log(x+1)-x+\mathrm{C} \): C is an arbitrary constant — Incorrect. It differs from \( (x+1)\log(x+1)-x+\mathrm{C} \) by \( 2\log(x+1) \), which is not a constant.
(D) Its particular solution is \( y=(x-1)\log(x+1)-x+5;y(0)=5 \) — Incorrect. It gives y(0) = 5, but it differs from the particular solution \( (x+1)\log(x+1)-x+5 \) by \( 2\log(x+1) \).
Tip: \( \int\log_e(x+1)\,dx=(x+1)\log_e(x+1)-x+\mathrm{C} \) is worth remembering; it is \( \int\log t\,dt=t\log t-t \) with t = x + 1.