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

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal: $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This

Vedclass · Accepted Answer

$y = e^{\frac{x}{y}}$

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x + y}$.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 $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = \frac{1}{y}$.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 $x=0, y=1$: $\frac{0}{1} = \ln(1) + C \implies C = 0$.Thus, $\frac{x}{y} = \ln y$, which implies $y = e^{\frac{x}{y}}$.

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

The general solution of the differential equation $\frac{d^2 y}{d x^2}+8 \frac{d y}{d x}+16 y=0$ is

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ is equal to

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$ for $-1 < x < 1$ such that $f(0)=0$. If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$,then $\alpha^2$ is equal to . . . . . .

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:

Let the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x \in(-1,1)$ pass through the origin. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module