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

A first or second order ODE in \(y(x)\) is called linear when the dependent variable \(y\) and all of its derivatives appear only to the first power, with no products of \(y\) and its derivatives, and no \(y\) buried inside a reciprocal or another nonlinear function. Check each option against that rule.

1. \((x+1)y' - y = e^x(x+1)^2\): linear. Divide by \((x+1)\) to get \(y' - \dfrac{1}{x+1}y = e^x(x+1)\), which is the standard linear form \(y' + P(x)y = Q(x)\) with coefficients depending only on \(x\).
2. \(y' - \dfrac{dx}{dy} = \dfrac{y}{x} - \dfrac{x}{y}\): not linear. Writing \(\dfrac{dx}{dy} = \dfrac{1}{y'}\) brings in a \(1/y'\) term, and the right side has \(y\) sitting in a denominator as \(x/y\); both break the linearity rule.
3. \(y'' + n^2x = 0\): linear. Rewritten as \(y'' = -n^2x\), this has \(y''\) with coefficient 1 and a forcing term that depends only on the independent variable \(x\); it is even directly integrable twice.
4. \(xy\,y' = 1 + x + y + xy\): not linear. The coefficient multiplying \(y'\) is \(xy\), which itself contains \(y\), so this is a product of the dependent variable and its own derivative, a hallmark of a nonlinear ODE.

Only the equations in (A) and (C) satisfy the linear ODE definition, so the correct selection is A and C.