Which of the following is true for y(x) that | vedclass

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

Question

(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.Th

Vedclass · Accepted Answer

$y(1) = e^{-\frac{1}{2}} - 1$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.This is a variable separable differential equation: $\frac{dy}{y + 1} = (x - 1) dx$.Integrating both sides: $\int \frac{dy}{y + 1} = \int (x - 1) dx$.$\ln|y + 1| = \frac{x^2}{2} - x + C$.Using the initial condition $y(0) = 0$: $\ln|0 + 1| = \frac{0^2}{2} - 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.Thus,$\ln|y + 1| = \frac{x^2}{2} - x$,which implies $y + 1 = e^{\frac{x^2}{2} - x}$.So,$y(x) = e^{\frac{x^2}{2} - x} - 1$.To find $y(1)$,substitute $x = 1$: $y(1) = e^{\frac{1^2}{2} - 1} - 1 = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1$.

Which of the following is true for $y(x)$ that satisfies the differential equation $\frac{dy}{dx} = xy - 1 + x - y$ with the initial condition $y(0) = 0$?

Similar Questions

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

The solution of the differential equation $3e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0$ is

Find the general solution of the differential equation: $y \log y \, dx - x \, dy = 0$.

The particular solution of $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ when $x=2, y=1$ is

Find a particular solution of the differential equation $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$,given that $y = 0$ when $x = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module