The equation of the curve passing through the | vedclass

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

Question

(A) Given the slope of the tangent $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal, $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This i

Vedclass · Accepted Answer

$x = y \ln(y)$

Vedclass · Answer

(A) Given the slope of the tangent $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal, $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This is a linear differential equation of the form $\frac{dx}{dy} - \frac{1}{y}x = 1$.The integrating factor $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.Multiplying by $IF$, we get $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = \frac{1}{y}$.Integrating both sides with respect to $y$, $\frac{x}{y} = \int \frac{1}{y} dy = \ln|y| + C$.Since the curve passes through $(0, 1)$, substitute $x=0$ and $y=1$: $\frac{0}{1} = \ln(1) + C \implies 0 = 0 + C \implies C = 0$.Thus, $\frac{x}{y} = \ln y$, which gives $x = y \ln y$.

The equation of the curve passing through the point $(0, 1)$ and having a slope of the tangent at any point $(x, y)$ equal to $\frac{y}{x + y}$ is:

Similar Questions

Find the particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$ $(x \neq 0)$,given that $y = 0$ when $x = \frac{\pi}{2}$.

If $f: R \rightarrow R$ is a differentiable function such that $f^{\prime}(x) > 2f(x)$ for all $x \in R$,and $f(0) = 1$,then:

The integrating factor of the differential equation $\frac{dy}{dx}(x \log x) + y = 2 \log x$ is given by

The solution of the differential equation $(x + 2y^3) \frac{dy}{dx} - y = 0$ is

The integrating factor of $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module