Question

A first-order ordinary differential equation is given as follows: \[\frac{dy}{dx} + x^2 y = 0\] Which ONE of the following options CORRECTLY represents the characteristics of this equation?

• A. Linear, homogeneous, and exact
• B. Nonlinear, nonhomogeneous, and exact
• C. Linear, homogeneous, and non-exact
• D. Nonlinear, nonhomogeneous, and non-exact

Hint:

Check the power of y and its derivative for linearity, check whether Q(x)=0 for homogeneity, and test partial M_y = N_x after writing it as M dx + N dy = 0 for exactness.

Answer 1

The Correct Option is C

Step 1: Check linearity. The equation \(\dfrac{dy}{dx}+x^2y=0\) can be written as \(\dfrac{dy}{dx}+P(x)y=Q(x)\) with \(P(x)=x^2\) and \(Q(x)=0\). Since \(y\) and its derivative appear only to the first power and are not multiplied together, the equation is linear.

Step 2: Check homogeneity. A first-order linear equation \(\dfrac{dy}{dx}+P(x)y=Q(x)\) is called homogeneous when \(Q(x)=0\). Here \(Q(x)=0\), so the equation is homogeneous.

Step 3: Check exactness. Write the equation in the differential form \(M\,dx+N\,dy=0\): rearranging \(dy+x^2y\,dx=0\) gives \(M=x^2y\) and \(N=1\). An equation is exact only if \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\). Here \(\dfrac{\partial M}{\partial y}=x^2\) and \(\dfrac{\partial N}{\partial x}=0\). Since \(x^2 \ne 0\) in general, the exactness condition fails, so the equation as written is NOT exact.

Step 4: Eliminate the wrong options. Options (A) and (B) wrongly call the equation exact, and options (B) and (D) wrongly call it nonlinear/nonhomogeneous even though \(y\) appears linearly and \(Q(x)=0\).

Step 5: Conclude. The equation is linear, homogeneous, and non-exact, which matches option (C).