Mathematics

Mathematics

Problem

If yy is twice differentiable function of xx, then the expression (1−x2)⋅d2ydx2−xdydx+y by means of the\text{expression } (1-x^2) \cdot \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y \text{ by means of the} transformation x=sin⁡t in terms of t is d2ydt2+λy.\text{transformation } x = \sin t \text{ in terms of } t \text{ is } \frac{d^2 y}{dt^2} + \lambda y. Thus, λ\lambda is .....
Intuitive explanation

This problem asks you to transform a differential equation using a substitution, which changes how derivatives behave. When you substitute x=sin⁡tx = \sin t, the chain rule creates relationships between ddx\frac{d}{dx} and ddt\frac{d}{dt} that simplify the original expression. The key is carefully applying the chain rule twice to rewrite dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2} in terms of derivatives with respect to tt, then substitute everything back into the given expression to reveal what λ\lambda must be.

Intuitive explanation

This problem asks you to transform a differential equation using a substitution, which changes how derivatives behave. When you substitute x=sin⁡tx = \sin t, the chain rule creates relationships between ddx\frac{d}{dx} and ddt\frac{d}{dt} that simplify the original expression. The key is carefully applying the chain rule twice to rewrite dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2} in terms of derivatives with respect to tt, then substitute everything back into the given expression to reveal what λ\lambda must be.

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