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

(C) The given differential equation is $(x^2-3y^2)dx+3xydy=0$.This can be rewritten as $\frac{dy}{dx} = \frac{3y^2-x^2}{3xy} = \frac{y}{x} - \frac{1}{

Vedclass · Accepted Answer

$2e^2$

Vedclass · Answer

(C) The given differential equation is $(x^2-3y^2)dx+3xydy=0$.This can be rewritten as $\frac{dy}{dx} = \frac{3y^2-x^2}{3xy} = \frac{y}{x} - \frac{1}{3}\frac{x}{y}$.Let $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting this into the equation: $v + x\frac{dv}{dx} = v - \frac{1}{3v}$.This simplifies to $x\frac{dv}{dx} = -\frac{1}{3v}$,or $3vdv = -\frac{dx}{x}$.Integrating both sides: $\int 3vdv = -\int \frac{dx}{x} \Rightarrow \frac{3v^2}{2} = -\ln|x| + C$.Substituting $v = \frac{y}{x}$: $\frac{3y^2}{2x^2} = -\ln|x| + C$.Given $y(1)=1$,we have $\frac{3(1)^2}{2(1)^2} = -\ln(1) + C \Rightarrow C = \frac{3}{2}$.Thus,$\frac{3y^2}{2x^2} = -\ln|x| + \frac{3}{2} \Rightarrow 3y^2 = 2x^2(\frac{3}{2} - \ln|x|) = 3x^2 - 2x^2\ln|x|$.At $x=e$,$3y^2(e) = 3e^2 - 2e^2\ln(e) = 3e^2 - 2e^2 = e^2$.Therefore,$6y^2(e) = 2(3y^2(e)) = 2e^2$.

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xydy=0$ with $y(1)=1$. Then $6y^2(e)$ is equal to $......$

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition $y(1) = 1$ is:

If a curve $y=f(x),$ passing through the point $(1,2),$ is the solution of the differential equation $2 x^{2} dy=\left(2 xy+y^{2}\right) dx,$ then $f\left(\frac{1}{2}\right)$ is equal to

The solution of the differential equation $2xy \frac{dy}{dx} = x^2 + 3y^2$ is (where $p$ is a constant):

The general solution of $y^2 dx + (x^2 - xy + y^2) dy = 0$ is:

The solution of the differential equation $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module