The slope of the tangent at (x, y) to a curve | vedclass

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

Question

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}

Vedclass · Accepted Answer

$2(x^2 - y^2) = 3x$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v}$.$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.Separating variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.Integrating both sides: $-\ln|1 - v^2| = \ln|x| + \ln|c|$.$-\ln|1 - \frac{y^2}{x^2}| = \ln|xc| \implies \ln|\frac{x^2}{x^2 - y^2}| = \ln|xc|$.$\frac{x^2}{x^2 - y^2} = xc \implies x = c(x^2 - y^2)$.Since the curve passes through $(2, 1)$,we have $2 = c(2^2 - 1^2) = c(3) \implies c = \frac{2}{3}$.Thus,$x = \frac{2}{3}(x^2 - y^2)$,which simplifies to $3x = 2(x^2 - y^2)$.

The slope of the tangent at $(x, y)$ to a curve passing through a point $(2, 1)$ is $\frac{x^2 + y^2}{2xy}$,then the equation of the curve is

Similar Questions

If $\sin \left(\frac{y}{x}\right)=\log |x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$,then $\alpha^2$ is equal to

Show that the differential equation $(x-y) \frac{dy}{dx} = x+2y$ is homogeneous and solve it.

Show that the differential equation $2 y e^{\frac{x}{y}} dx + (y - 2 x e^{\frac{x}{y}}) dy = 0$ is homogeneous and find its particular solution,given that $x = 0$ when $y = 1$.

In a single throw of two dice,what is the probability of obtaining a sum of numbers greater than $7$,given that $4$ appears on the first die?

Let $f(x) = \sqrt{\lim_{r \rightarrow x} \left\{ \frac{2r^2 \left[(f(r))^2 - f(x)f(r)\right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\}}$ be differentiable in $(-\infty, 0) \cup (0, \infty)$ and $f(1) = 1$. Then the value of $ea$,such that $f(a) = 0$,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module