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

(C) Given the differential equation for the slope of the tangent:$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right)$This is a homogeneous d

Vedclass · Accepted Answer

$y=x 	an ^{-1}\left[\log \left(\frac{e}{x}ight)ight]$

Vedclass · Answer

(C) Given the differential equation for the slope of the tangent:$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}ight)$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} = v - \cos^2(v)$$x\frac{dv}{dx} = -\cos^2(v)$Separating the variables:$\sec^2(v) dv = -\frac{1}{x} dx$Integrating both sides:$\int \sec^2(v) dv = -\int \frac{1}{x} dx$$	an(v) = -\log|x| + C$Substituting $v = \frac{y}{x}$ back:$	an\left(\frac{y}{x}ight) = -\log|x| + C$The curve passes through $\left(1, \frac{\pi}{4}ight)$,so:$	an\left(\frac{\pi/4}{1}ight) = -\log(1) + C$$	an\left(\frac{\pi}{4}ight) = 0 + C \Rightarrow C = 1$Thus,$	an\left(\frac{y}{x}ight) = 1 - \log(x) = \log(e) - \log(x) = \log\left(\frac{e}{x}ight)$Therefore,$y = x 	an^{-1}\left[\log\left(\frac{e}{x}ight)ight]$.

The slope of the tangent at $(x, y)$ to a curve passing through $\left(1, \frac{\pi}{4}\right)$ is given by $\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)$,then the equation of the curve is

Similar Questions

The general solution of the differential equation $x \cos \frac{y}{x}(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$ is

Suppose that $f(x, y)$ and $g(x, y)$ are homogeneous functions of the same order. If $x=Vy$ reduces the equation $\frac{dy}{dx}=\frac{f(x, y)}{g(x, y)}$ to the form $\frac{dV}{dy}=\frac{1}{y}(F(V))$,then $F(V)=$

The general solution of the differential equation $(3y - 7x + 7)dx + (7y - 3x + 3)dy = 0$ is

If the solution curve $y=y(x)$ of the differential equation $y^{2} dx + (x^{2} - xy + y^{2}) dy = 0$ passes through the point $(1, 1)$ and intersects the line $y = \sqrt{3}x$ at the point $(\alpha, \sqrt{3}\alpha)$,then the value of $\log_{e}(\sqrt{3}\alpha)$ is equal to

The general solution of the differential equation $(x+y) y dx + (y-x) x dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module