If $y = f(x)$ is a monotonically increasing function such that $(\frac{dy}{dx})^2 = 6-\frac{dy}{dx}$ and $y(0) = 5$, then $y(3) = \cdots$

Question

The solution of the differential equation $ x \cos y \, dy = (x e^x \log x + e^x) \, dx $ is:

Answer 1

Step 1: Recognise a constant slope:
The equation links $y'$ only to itself, so $y'$ is constant, being a root of the quadratic.

Step 2: Pick the root:
The roots of $t^2 + t - 6 = 0$ are 2 and -3. An increasing function has non-negative slope, so the slope is 2.

Step 3: Integrate:
$y = 2x + 5$, so $y(3) = 11$.

Final Answer:
The value is 11, option (D). \[ \boxed{11} \]

If \(y = f(x)\) is a monotonically increasing function such that \((\frac{dy}{dx})^2 = 6-\frac{dy}{dx}\) and \(y(0) = 5\), then \(y(3) = \cdots\)

Show Hint

The Correct Option is D

Solution and Explanation

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