If y = y(x) is the solution of the differenti | vedclass

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

Question

(B) Given the differential equation: $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$.Divide by $2x^{2}y^{2}$ to transform it into a Bernoulli equation or su

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Given the differential equation: $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$.Divide by $2x^{2}y^{2}$ to transform it into a Bernoulli equation or substitute $y = vx$:$\frac{dy}{dx} - \frac{y}{x} = -\frac{3}{2} \left(\frac{y}{x}\right)^{2}$.Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} - v = -\frac{3}{2} v^{2}$.$x \frac{dv}{dx} = -\frac{3}{2} v^{2}$.Separating variables: $\frac{dv}{v^{2}} = -\frac{3}{2} \frac{dx}{x}$.Integrating both sides: $-\frac{1}{v} = -\frac{3}{2} \ln|x| + C$.Substituting $v = \frac{y}{x}$ back: $-\frac{x}{y} = -\frac{3}{2} \ln|x| + C$.Using the condition $y(e) = \frac{e}{3}$: $-\frac{e}{e/3} = -\frac{3}{2} \ln(e) + C \implies -3 = -\frac{3}{2} + C \implies C = -\frac{3}{2}$.So,$-\frac{x}{y} = -\frac{3}{2} \ln|x| - \frac{3}{2}$.For $x = 1$: $-\frac{1}{y} = -\frac{3}{2} \ln(1) - \frac{3}{2} \implies -\frac{1}{y} = -\frac{3}{2} \implies y = \frac{2}{3}$.

If $y = y(x)$ is the solution of the differential equation $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$ such that $y(e) = \frac{e}{3}$,then $y(1)$ is equal to

Similar Questions

The solution of the differential equation $y \ dx - x \ dy = xy \ dx$ is . . . . . .

The solution curve of the differential equation,$(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$,which passes through the point $(0,1)$,is

The solution of the differential equation $\sin ^{-1}\left(\frac{dy}{d x}\right)=x+y$ is

The general solution of $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ is

The general solution of the differential equation $(x-y)^2 \frac{dy}{dx} = a^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module