#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 73 textbook solution.

Answer:  $-1+2e^{x^{2}/2}$

Given: The slope of the tangent to the curve at any point $(x , y)$ is equal to the sum of x coordinate and the product of x and y coordinate of that point

Hint: You must know about integrating factor

Explanation: Slope of the tangent to the curve at $(x , y)=\frac{dy}{dx}$

Given that,

Slope of the tangent to the curve at point  $(x , y)$ is equal to the sum of x coordinate and the product of x and y coordinate of that point

So our equation becomes,

\begin{aligned} &\frac{d y}{d x}=x+x y \\ \\&\Rightarrow \frac{d y}{d x}-x y=x \end{aligned}

Differential equation is of the form

\begin{aligned} &\frac{d y}{d x}+Py=Q \\ \end{aligned}

Where

\begin{aligned} &\mathrm{P}=-\mathrm{x} \text { or }\\ \\&\theta=x\\ \end{aligned}

If

\begin{aligned} &=e^{P d \int x}\\ \\&=e^{-\int x d x}\\ \\&=e^{-x^{2} / 2} \end{aligned}

Solution is

\begin{aligned} &\mathrm{y} \cdot(\mathrm{I} \cdot \mathrm{f} \cdot)=\int(\theta \times \mathrm{I} \cdot \mathrm{f} \cdot) \mathrm{d} \mathrm{x}+\mathrm{c} \\ &\mathrm{ye}^{-\frac{\mathrm{x}^{2}}{2}}=\int \mathrm{xe}^{-\frac{\mathrm{x}^{2}}{2}} \mathrm{dx}+\mathrm{c} \end{aligned}

Putting ,

\begin{aligned} &-\frac{x^{2}}{2}=t \\ \\&\Rightarrow \frac{-2 x}{2} d x=d t \\ \\&\Rightarrow x d x=a b \end{aligned}

Thus our equation becomes

\begin{aligned} &y e^{-x^{2} / 2}=\int-e^{t} d t+c\\ &y e^{-x^{2} / 2}=-e^{t}+c\\ &\text { Putting back } t=-\frac{x^{2}}{2}\\ &\Rightarrow \mathrm{ye}^{-\frac{\mathrm{x}^{2}}{2}}\\ &=-\mathrm{e}^{-\mathrm{x} / 2}+\mathrm{c}\\ &\Rightarrow \frac{y}{\mathrm{e}^{\mathrm{x}^{2} / 2}} \end{aligned}

\begin{aligned} &=-\frac{1}{e^{x^{2} / 2}}+c \\ &\Rightarrow y=-1+c e^{x^{2 / 2}}-(1) \end{aligned}

Since curve passes through (0,1)

Putting

\begin{aligned} &\mathrm{x}=0, \\ \\&\mathrm{y}=1 \mathrm{in}(1) \\ \\&\mathrm{y}=-1+\mathrm{ce}^{\mathrm{x}^{2} / 2} \\ \\&1=-1+\mathrm{ce}^{\mathrm{a}_{2} / 2} \\ \\&\Rightarrow 1+1=\mathrm{C} \end{aligned}                                  $[\because e^{0}=1]$

$\Rightarrow c=2$

Putting value of c in (1)

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

$\therefore$ Equation of the curve  is  $-1+2e^{x^{2}/2}$