Question

Which of the following ordinary differential equations is/are linear?

• A. \((x+1)\dfrac{dy}{dx} - y = e^x(x+1)^2\)
• B. \(\dfrac{dy}{dx} - \dfrac{dx}{dy} = \dfrac{y}{x} - \dfrac{x}{y}\)
• C. \(\dfrac{d^2y}{dx^2} + n^2x = 0\), where \(n\) is a constant
• D. \(xy\dfrac{dy}{dx} = 1 + x + y + xy\)

Hint:

Check whether y and its derivative appear only to the first power, with coefficients that depend on x alone.

Answer 1

The Correct Option is A, C

Step 1: Recall the test for linearity.
An ODE is linear in $y$ if it can be arranged as $a_n(x) y^{(n)} + \dots + a_1(x) y' + a_0(x) y = f(x)$, meaning every term with $y$ or a derivative of $y$ is multiplied only by a function of $x$, never by $y$ itself, and $y$ never appears inside a reciprocal, square root or other nonlinear wrapper.

Step 2: Test option A.
$(x+1)y' - y = e^x(x+1)^2$ already has this shape: coefficient $(x+1)$ on $y'$, coefficient $-1$ on $y$, both functions of $x$ only. So A passes.

Step 3: Test option B.
$y' - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$ needs $\frac{dx}{dy} = 1/y'$, giving a $1/y'$ term, and it also has $x/y$ on the right, so $y$ appears in a denominator. Both features fail the test, so B is nonlinear.

Step 4: Test option C.
$y'' + n^2x = 0$ is just $y'' = -n^2x$: the only $y$-term is $y''$ with coefficient 1, and the rest is a pure function of $x$. This passes the test cleanly.

Step 5: Test option D.
$xy\,y' = 1+x+y+xy$ has $y'$ multiplied by $xy$, a term that includes $y$, so the coefficient of $y'$ is not a function of $x$ alone. This fails, so D is nonlinear.

Final Answer:
Options A and C meet the linear ODE definition and B and D do not. \[ \boxed{\text{A and C}} \]