Let y=y(x), y0,be a solution curve of the di | vedclass

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

Question

(A) Given the differential equation: $(1+x^2) dy = y(x-y) dx$.Dividing by $(1+x^2) dx$,we get: $\frac{dy}{dx} = \frac{xy - y^2}{1+x^2} = \frac{x}{1+x^

Vedclass · Accepted Answer

$e^{3\beta^{-1}} = e(3+2\sqrt{2})$

Vedclass · Answer

(A) Given the differential equation: $(1+x^2) dy = y(x-y) dx$.Dividing by $(1+x^2) dx$,we get: $\frac{dy}{dx} = \frac{xy - y^2}{1+x^2} = \frac{x}{1+x^2}y - \frac{1}{1+x^2}y^2$.This is a Bernoulli differential equation. Rearranging gives: $\frac{dy}{dx} - \frac{x}{1+x^2}y = -\frac{1}{1+x^2}y^2$.Divide by $y^2$: $y^{-2} \frac{dy}{dx} - \frac{x}{1+x^2}y^{-1} = -\frac{1}{1+x^2}$.Let $t = y^{-1}$,then $\frac{dt}{dx} = -y^{-2} \frac{dy}{dx}$.Substituting this into the equation: $-\frac{dt}{dx} - \frac{x}{1+x^2}t = -\frac{1}{1+x^2} \implies \frac{dt}{dx} + \frac{x}{1+x^2}t = \frac{1}{1+x^2}$.This is a linear differential equation. The integrating factor $I.F. = e^{\int \frac{x}{1+x^2} dx} = e^{\frac{1}{2} \ln(1+x^2)} = \sqrt{1+x^2}$.The solution is $t \cdot \sqrt{1+x^2} = \int \frac{1}{1+x^2} \cdot \sqrt{1+x^2} dx = \int \frac{1}{\sqrt{1+x^2}} dx = \ln(x + \sqrt{1+x^2}) + C$.Since $t = \frac{1}{y}$,we have $\frac{\sqrt{1+x^2}}{y} = \ln(x + \sqrt{1+x^2}) + C$.Given $y(0)=1$,we have $\frac{\sqrt{1}}{1} = \ln(0+1) + C \implies 1 = 0 + C \implies C=1$.So,$\frac{\sqrt{1+x^2}}{y} = \ln(x + \sqrt{1+x^2}) + 1 = \ln(x + \sqrt{1+x^2}) + \ln e = \ln(e(x + \sqrt{1+x^2}))$.For $x = 2\sqrt{2}$,$y = \beta$: $\frac{\sqrt{1+(2\sqrt{2})^2}}{\beta} = \ln(e(2\sqrt{2} + \sqrt{1+8})) = \ln(e(2\sqrt{2} + 3))$.$\frac{3}{\beta} = \ln(e(3+2\sqrt{2})) \implies e^{3\beta^{-1}} = e(3+2\sqrt{2})$.

Let $y=y(x), y>0$,be a solution curve of the differential equation $(1+x^2) dy = y(x-y) dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta$,then

Similar Questions

If $y=f(x)$ is the solution of the differential equation $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$ with $f(0)=0$, then $f(\frac{\pi}{3})=$

The solution of $\frac{dy}{dx} + p(x)y = 0$ is

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by $5$ is

The solution of the equation $\frac{dy}{dx} + 2y \tan x = \sin x$ satisfying $y = 0$ when $x = \frac{\pi}{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module