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

(A) The given differential equation is $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$.Rearranging the terms to separate the variables,we get:$\frac{y}

Vedclass · Accepted Answer

$1-4e^{6}$

Vedclass · Answer

(A) The given differential equation is $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$.Rearranging the terms to separate the variables,we get:$\frac{y}{\sqrt{1-y^{2}}} dy = -x e^{x} dx$.Integrating both sides:$\int \frac{y}{\sqrt{1-y^{2}}} dy = -\int x e^{x} dx$.Let $u = 1-y^{2}$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$. The left side becomes:$-\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} (2u^{1/2}) = -\sqrt{1-y^{2}}$.For the right side,using integration by parts $\int x e^{x} dx = x e^{x} - e^{x} + C = e^{x}(x-1) + C$.Thus,$-\sqrt{1-y^{2}} = -(e^{x}(x-1) + C) = -e^{x}(x-1) - C$.So,$\sqrt{1-y^{2}} = e^{x}(x-1) + C$.Using the condition $y(1) = -1$:$\sqrt{1-(-1)^{2}} = e^{1}(1-1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.Therefore,$\sqrt{1-y^{2}} = e^{x}(x-1)$.Squaring both sides,$1-y^{2} = e^{2x}(x-1)^{2}$.At $x=3$,$1-y^{2} = e^{6}(3-1)^{2} = 4e^{6}$.$y^{2} = 1 - 4e^{6}$.

Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$,with $y(1) = -1$. Then the value of $(y(3))^{2}$ is equal to:

Similar Questions

The equation of the curve passing through $(3, 9)$ which satisfies the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$ is

The solution of the differential equation ${x^4}\frac{{dy}}{{dx}} + {x^3}y + \text{cosec}(xy) = 0$ is equal to

Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$ when $x = 0$.

If the solution curve $y=y(x)$ of the differential equation $(1+y^2)(1+\log_e x) dx + x dy = 0, x>0$ passes through the point $(1,1)$ and $y(e) = \frac{\alpha-\tan(3/2)}{\beta+\tan(3/2)}$,then $\alpha+2\beta$ is equal to:

The solution of the equation $(1 + x^2)\frac{dy}{dx} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module