If y = f(x) is a monotonically increasing fun | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $p = \frac{dy}{dx}$. The given equation is $p^2 = 6 - p$, which rearranges to $p^2 + p - 6 = 0$.Factoring the quadratic, we get $(p + 3)(p - 2

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) Let $p = \frac{dy}{dx}$. The given equation is $p^2 = 6 - p$, which rearranges to $p^2 + p - 6 = 0$.Factoring the quadratic, we get $(p + 3)(p - 2) = 0$, so $p = 2$ or $p = -3$.Since $y = f(x)$ is a monotonically increasing function, $\frac{dy}{dx} \ge 0$, so we must have $p = 2$.Integrating $\frac{dy}{dx} = 2$ with respect to $x$, we get $y = 2x + C$.Using the initial condition $y(0) = 5$, we find $5 = 2(0) + C$, so $C = 5$.Thus, the function is $y = 2x + 5$.Evaluating at $x = 3$, we get $y(3) = 2(3) + 5 = 6 + 5 = 11$.

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

Similar Questions

The differential equation $x^2(y+1) dx + y^2(x-1) dy = 0$ has the general solution given by (where $C$ is a constant of integration.)

The particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ with the initial condition $x = 0, y = 1$ is . . . . . . .

The solution of the differential equation $x \sec y \frac{dy}{dx} = 1$ is

Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y \, dx + \cos x \sin y \, dy = 0$.

The solution of the equation $\frac{dy}{dx} = \frac{1}{x+y+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module