Let y=y_{1}(x) and y=y_{2}(x) be two distinct | vedclass

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

Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} - y = x$.The integrating factor $(IF)$ is $e^{\int -1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} - y = x$.The integrating factor $(IF)$ is $e^{\int -1 dx} = e^{-x}$.Multiplying both sides by $e^{-x}$,we get $\frac{d}{dx}(y e^{-x}) = x e^{-x}$.Integrating both sides,$y e^{-x} = \int x e^{-x} dx = -x e^{-x} - e^{-x} + C$.Thus,the general solution is $y = -x - 1 + C e^{x}$.For $y_{1}(0) = 0$,we have $0 = -0 - 1 + C(1) \Rightarrow C = 1$. So,$y_{1}(x) = e^{x} - x - 1$.For $y_{2}(0) = 1$,we have $1 = -0 - 1 + C(1) \Rightarrow C = 2$. So,$y_{2}(x) = 2e^{x} - x - 1$.To find the points of intersection,set $y_{1}(x) = y_{2}(x)$:$e^{x} - x - 1 = 2e^{x} - x - 1$.This simplifies to $e^{x} = 0$,which has no real solution for $x$.Therefore,the number of points of intersection is $0$.

Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be two distinct solutions of the differential equation $\frac{dy}{dx}=x+y$,with $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then,the number of points of intersection of $y=y_{1}(x)$ and $y=y_{2}(x)$ is.

Similar Questions

The solution of the differential equation $\sqrt{1-y^2} dx + x dy - \sin^{-1} y dy = 0$ is

$A$ function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0$ is such that $y \rightarrow 0$ as $x \rightarrow \infty$. Which of the following statements is correct?

The general solution of a differential equation of the type $\frac{dx}{dy} + P_{1}x = Q_{1}$ is

The solution of the differential equation $(y^{2}+2x) \frac{dy}{dx}=y$ satisfies $x=1, y=1$. Then the solution is

If $y=y(x)$ is the solution of the differential equation $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module