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

Option: 2
$1$

Option: 3
$2$

Option: 4
$3$

Answers (1)

best_answer

$\mathrm{\frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}-y=x}$

$\mathrm{If =e^{-x}}$

$\mathrm{\therefore solution\; is\; y \cdot e^{-x}=\int x \cdot e^{-x} d x}$

$\mathrm{\Rightarrow y e^{-x}=-x e^{-x}-e^{-x}+c} \\$

$\mathrm{\Rightarrow y=-x-1+c e^{x}} \\$

$\mathrm{y_{1}(0)=0 \Rightarrow c=1} \\$

$\mathrm{\therefore y_{1}=-x-1+e^{x}}$ ............(1)

$\mathrm{y_{2}(0)=1 \Rightarrow c=2} \\$

$\mathrm{\therefore y_{2}=-x-1+2 e^{x} \ldots \text { (2) }} \\$

$\mathrm{\text { Now } y_{2}-y_{1}=e^{x}>0 \quad \therefore y_{2} \neq y_{1}}$

$\therefore$ Number of point of intersections of $\mathrm{y_{1}\& \; y_{2}}$ is zero.

Hence correct option is 1

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