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

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

Question

(B) Given the differential equation $\frac{dy}{dx}=(x+y+2)^2$ with $y(0)=-2$.Let $v=x+y+2$,then $\frac{dv}{dx}=1+\frac{dy}{dx}$.Substituting into the

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) Given the differential equation $\frac{dy}{dx}=(x+y+2)^2$ with $y(0)=-2$.Let $v=x+y+2$,then $\frac{dv}{dx}=1+\frac{dy}{dx}$.Substituting into the equation: $\frac{dv}{dx}-1=v^2 \Rightarrow \frac{dv}{dx}=1+v^2$.Integrating both sides: $\int \frac{dv}{1+v^2} = \int dx \Rightarrow 	an^{-1}(v) = x+C$.Thus,$	an^{-1}(x+y+2) = x+C$.At $x=0, y=-2$,we have $	an^{-1}(0-2+2) = 0+C \Rightarrow C=0$.So,$x+y+2 = 	an(x) \Rightarrow y = 	an(x)-x-2$.For $x \in [0, \frac{\pi}{3}]$,$f'(x) = \sec^2(x)-1 = 	an^2(x) \ge 0$,so $f(x)$ is increasing.Minimum value $\beta = f(0) = 	an(0)-0-2 = -2$.Maximum value $\alpha = f(\frac{\pi}{3}) = 	an(\frac{\pi}{3})-\frac{\pi}{3}-2 = \sqrt{3}-\frac{\pi}{3}-2$.Now,$(3\alpha+\pi)^2+\beta^2 = (3(\sqrt{3}-\frac{\pi}{3}-2)+\pi)^2+(-2)^2$.$= (3\sqrt{3}-\pi-6+\pi)^2+4 = (3\sqrt{3}-6)^2+4$.$= (27+36-36\sqrt{3})+4 = 67-36\sqrt{3}$.Comparing with $\gamma+\delta\sqrt{3}$,we get $\gamma=67$ and $\delta=-36$.Therefore,$\gamma+\delta = 67-36 = 31$.

Similar Questions

Solve the differential equation $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)\,dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y\,dy = 0$.

Verify that the given function $y = x \sin x$ is a solution of the differential equation $x y^{\prime} = y + x \sqrt{x^2 - y^2}$ (where $x \neq 0$ and $x > y$ or $x < -y$).

Let $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$,where $a, b, c$ are constants,represent a circle passing through the point $(2, 5)$. Then the shortest distance of the point $(11, 6)$ from this circle is

$A$ function $y = f(x)$ satisfies the condition $f'(x) \sin x + f(x) \cos x = 1$,where $f(x)$ is bounded as $x \rightarrow 0$. If $I = \int_{0}^{\frac{\pi}{2}} f(x) \, dx$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module