Question

Which of the following differential equations has y = x as one of its particular solution?

Options

• \[\frac{d^2 y}{d x^2} - x^2 \frac{dy}{dx} + xy = x\]

• \[\frac{d^2 y}{d x^2} + x\frac{dy}{dx} + xy = x\]

• \[\frac{d^2 y}{d x^2} - x^2 \frac{dy}{dx} + xy = 0\]

• \[\frac{d^2 y}{d x^2} + x\frac{dy}{dx} + xy = 0\]

Answer 1

\[\frac{d^2 y}{d x^2} - x^2 \frac{dy}{dx} + xy = 0\]

 

We have,

y = x           .....(1)

Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dx} = 1 . . . . . \left( 2 \right)\]
Differentiating again with respect to x, we get
\[ \Rightarrow \frac{d^2 y}{d x^2} = 0\]
\[ \Rightarrow \frac{d^2 y}{d x^2} + x^2 = x^2 \]
\[ \Rightarrow \frac{d^2 y}{d x^2} + x \times x = x^2 \times 1\]
\[ \Rightarrow \frac{d^2 y}{d x^2} + xy = x^2 \times 1 ............\left[\text{Using }\left( 1 \right) \right]\]
\[ \Rightarrow \frac{d^2 y}{d x^2} + xy = x^2 \frac{dy}{dx} .............\left[ \text{Using }\left( 2 \right) \right]\]
\[ \Rightarrow \frac{d^2 y}{d x^2} - x^2 \frac{dy}{dx} + xy = 0\]