Question

we are given that \\( \frac { d y } { d x } = x ^ { 2 } - 2 y \\).
find an expression for \\( \frac { d ^ { 2 } y } { d x ^ { 2 } } \\) in terms of \\( x \\) and \\( y \\).
\\( \frac { d ^ { 2 } y } { d x ^ { 2 } } = \\)

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(\frac{dy}{dx}=x^{2}-2y\) with respect to \(x\).
Using the sum - difference rule \((u - v)^\prime=u^\prime - v^\prime\) (where \(u = x^{2}\) and \(v = 2y\)), and the power rule \((x^{n})^\prime=nx^{n - 1}\) (\(n = 2\) for \(x^{2}\)) and the chain rule \((y)^\prime=\frac{dy}{dx}\).
\(\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(x^{2})-\frac{d}{dx}(2y)\)
\(\frac{d^{2}y}{dx^{2}} = 2x-2\frac{dy}{dx}\)

Step2: Substitute \(\frac{dy}{dx}\)

Since \(\frac{dy}{dx}=x^{2}-2y\), substitute it into the above - equation.
\(\frac{d^{2}y}{dx^{2}}=2x - 2(x^{2}-2y)\)

Step3: Simplify the expression

Expand the right - hand side: \(\frac{d^{2}y}{dx^{2}}=2x-2x^{2}+4y\)

Answer:

\(2x - 2x^{2}+4y\)