Let the curve y = y(x) be the solution of the | vedclass

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

Question

(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.Integrating both sides with respect to $x$,we get:$y(x) = \int 2(x + 1) dx = (x + 1)^2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.Integrating both sides with respect to $x$,we get:$y(x) = \int 2(x + 1) dx = (x + 1)^2 + C = x^2 + 2x + 1 + C$.Let $K = 1 + C$,so $y(x) = (x + 1)^2 + C$.The curve is a parabola opening upwards with vertex at $(-1, C)$. For the curve to bound an area with the $x-$axis,it must lie below the $x-$axis,so $C  0$.The roots of $y(x) = 0$ are $(x + 1)^2 = -C$,so $x = -1 \pm \sqrt{-C}$.The area bounded by the curve and the $x-$axis is given by:$A = \int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} (0 - ((x + 1)^2 + C)) dx = -\int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} ((x + 1)^2 + C) dx$.Let $u = x + 1$,then $du = dx$. The limits change to $-\sqrt{-C}$ to $\sqrt{-C}$.$A = -\int_{-\sqrt{-C}}^{\sqrt{-C}} (u^2 + C) du = -[\frac{u^3}{3} + Cu]_{-\sqrt{-C}}^{\sqrt{-C}} = -[(\frac{(-C)^{3/2}}{3} + C\sqrt{-C}) - (\frac{-(-C)^{3/2}}{3} - C\sqrt{-C})]$.$A = -[\frac{2}{3}(-C)^{3/2} + 2C\sqrt{-C}] = -[\frac{2}{3}(-C)\sqrt{-C} - 2(-C)\sqrt{-C}] = -[-\frac{4}{3}(-C)^{3/2}] = \frac{4}{3}(-C)^{3/2}$.Given $A = \frac{4\sqrt{8}}{3} = \frac{4(2\sqrt{2})}{3} = \frac{8\sqrt{2}}{3}$.So,$\frac{4}{3}(-C)^{3/2} = \frac{8\sqrt{2}}{3} \Rightarrow (-C)^{3/2} = 2\sqrt{2} = (\sqrt{2})^3$.Thus,$-C = 2$,which means $C = -2$.The function is $y(x) = (x + 1)^2 - 2 = x^2 + 2x - 1$.Therefore,$y(1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2$.

Let the curve $y = y(x)$ be the solution of the differential equation,$\frac{dy}{dx} = 2(x + 1)$. If the numerical value of the area bounded by the curve $y = y(x)$ and the $x-$axis is $\frac{4\sqrt{8}}{3}$,then the value of $y(1)$ is equal to

Similar Questions

The population of a city increases at the rate of $3 \%$ per year. If the population is $p$ at time $t$,then the equation of $p$ in terms of $t$ is . . . . . . .

The differential equation $x \frac{dy}{dx} = 2y$ represents ....

The area bounded by a curve,the $x$-axis,and the ordinate of some point $(x, y)$ of the curve is equal to the length of the corresponding arc of the curve from a fixed point. If the curve passes through the point $P(0, 1)$,then the equation of this curve is:

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

If a body cools from $135^{\circ} C$ to $80^{\circ} C$ at a room temperature of $25^{\circ} C$ in $60 \text{ minutes}$,then the temperature of the body after $2 \text{ hours}$ is: (in $^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module