#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 74 textbook solution.

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

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:  We know that

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

According to the given

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

This is of the form

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

Where

\begin{aligned} &\mathrm{P}=-\mathrm{1} \text { and }\\ \\&\theta=x\\ \end{aligned}

Finding integrating factor

If,

\begin{aligned} &=e^{\int P d x} \\ \\&=e^{\int(-1) d x} \\ \\&=e^{-x d x} \end{aligned}                                                          $\left [ \because \int 1\; dx =x+c\right ]$

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}^{-\mathrm{x}}=\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \end{aligned}

Using integration by parts,

\begin{aligned} &\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}=\mathrm{x} \cdot \frac{\mathrm{e}^{-\mathrm{x}}}{-1}-\int(1) \frac{\mathrm{e}^{-\mathrm{x}}}{-1} \mathrm{dx} \\ &=-\mathrm{xe}^{-\mathrm{x}}+\frac{\mathrm{e}^{-\mathrm{x}}}{-1}+\mathrm{c} \\ &=-\mathrm{xe}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{C} \end{aligned}

Put in (1)

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

Divide by $e^{-x}$

$\Rightarrow y=-x-1+c e^{x}-(2)$

Since curve passes through origin,

Putting x=0  and y=0  in (2)

$\begin{gathered} 0 \quad=0-1+\mathrm{Ce}^{0} \\ \Rightarrow \mathrm{c}=1 \end{gathered}$                                   $[\because e^{0}=1]$

Put vaue of c in (2)

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