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### Answers (1)

Answer:$y+1=2 e^{\frac{x^{2}}{2}}$

Given: The slope of tangent at each point of a curve is equal to the sum of abscissa and product of abscissa and ordinate of the point i.e. $\frac{d y}{d x}=x+y$

To find: The curves that pass through the point $\left ( 0,1 \right )$

Hint: Use linear differential equation to find the equation of the curve.

Solution: we have  $\frac{d y}{d x}=x+y$

$=\frac{d y}{d x}-x y=x \ldots(i)$

This is a linear differential equation of the type

$=\frac{d y}{d x}-P y=Q$

Where $P=-x, Q=x$

\begin{aligned} &=I f=e^{\int p d x} \\\\ &=I f=e^{-\int x d x} \\\\ &=I f=e^{-\frac{x^{2}}{2}} \end{aligned}

So solution of equation is given by

$=y e^{-\frac{x^{2}}{2}}=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x+C \ldots(i i)$

\begin{aligned} &\text { Let } I=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x \\\\ &\text { Let }-\frac{x^{2}}{2}=t \end{aligned}

Differentiating on both sides with respect to x

\begin{aligned} &=-\frac{2 x}{2} d x=d t \\\\ &=-x d x=d t \\\\ &=x d x=-d t \end{aligned}

\begin{aligned} &=I=\int e^{-t} d t \\\\ &=I=-e^{-t} \\\\ &=I=-e^{-\frac{x^{2}}{2}} \end{aligned}

Substituting value of $I$ in equation (ii) we get

$=y\left(e^{-\frac{x^{2}}{2}}\right)=-e^{-\frac{x^{2}}{2}}+C$

Dividing by $e^{-\frac{x^{2}}{2}}$

$=y=-1+C e^{\frac{x^{2}}{2}} \ldots(i i i)$

As curve passes through $\left ( 0,1 \right )$

\begin{aligned} &=1=-1+C e^{0} \\\\ &=C=2 \end{aligned}

Substituting value of C in equation (iii) we get

\begin{aligned} &=y=-1+2 e^{\frac{x^{2}}{2}} \\\\ &=y+1=2 e^{\frac{x^{2}}{2}} \end{aligned}

Hence, equation of required curve is found.

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