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

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(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

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$0$

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(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.

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