#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 75 textbook solution.

Answer:  $y= \begin{cases}4-x-2 e^{x} & \text { iff } \frac{d y}{d x}=+(x+y-5) \\ 6-x-4 e^{-x} & \text { iff } \frac{d y}{d x}=-(x+4-5)\end{cases}$

Given: Sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5

Hint: Using integration by parts and integration factor

Explanation: We know that

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

According to the given

\begin{aligned} &\left|\frac{d y}{d x}\right|+5=x+y \\ &\Rightarrow\left|\frac{d y}{d x}\right|=x+y-5 \\ &\Rightarrow \frac{d y}{d x}=\pm(x+y-5) \end{aligned}

So we will take both (+) ve and (-) ve sign and then solve it

Taking (+) ve sign:

\begin{aligned} &\frac{d y}{d x}=x+y-5 \\ &\Rightarrow \frac{d y}{d x}-y=x-5 \end{aligned}

Equation is of the form

$\frac{d y}{d x}+P_{y}=Q$

Where,

\begin{aligned} &P=-1 \\ &\theta=x-5 \\ &I \cdot F \cdot=e^{\int P d x} \\ &\qquad=e^{\int(-1) d x} \\ &\; \; \; \; \quad=e^{-x} \end{aligned}

Solution is

\begin{aligned} &\mathrm{y}(\mathrm{I} \cdot \mathrm{F})=\int(\theta \times \mathrm{I} \cdot \mathrm{f} \cdot) \mathrm{d} \mathrm{x}+\mathrm{c} \\ &\mathrm{ye}^{-\mathrm{x}}=\int(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\int \quad(1) \frac{\mathrm{e}^{-\mathrm{x}}}{-1} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}+\int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c} \end{aligned}

Divided by $e^{-x}$

\begin{aligned} &\Rightarrow \mathrm{y}=-(\mathrm{x}-5)-1+\mathrm{ce}^{\mathrm{x}}-(1) \\ \\&\Rightarrow \mathrm{y}_{\mathrm{y}}=5-\mathrm{x}-1+\mathrm{ce}^{\mathrm{x}} \\ \\&\; \; \; \; \; \; \quad=4-\mathrm{x}+\mathrm{ce}^{\mathrm{x}} \end{aligned}

Since the curve passes through -the point (0,2)

Put x=0 and

$y=2$

\begin{aligned} &2=4-0+c e^{\circ} \\ \\&\Rightarrow 2=4+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \mathbf{e}^{0}=1\right] \\ \\&\Rightarrow C=-2 \end{aligned}

Equation of the curve is

$y=4-x-2 e^{x}$

Taking (-) ve sign:

\begin{aligned} &\frac{d y}{d x}=-x-y+5 \\ &\Rightarrow \frac{d y}{d x}+y=-x+5 \end{aligned}

Equation is of the form

$\frac{d y}{d x}+P_{y}=Q$

Where,

\begin{aligned} &\mathrm{P}=1 \\ &\theta=-\mathrm{x}+5 \\ &\mathrm{I} \cdot \mathrm{F} \cdot=\mathrm{e}^{\int \mathrm{Pdx}} \\ &\qquad \begin{aligned} &=\mathrm{e}^{\int(1) \mathrm{dx}} \\ &=\mathrm{e}^{\mathrm{x}} \end{aligned} \end{aligned}                              $[\because \int 1dx=x+c]$

Solution is

\begin{aligned} &y(I \cdot F)=\int(\theta \times I \cdot f \cdot) d x+c \\ &\Rightarrow y e^{x}=(5-x) e^{-x}-\int(-1) e^{x} d x+c \end{aligned}                                      [Using integration by parts]

$\Rightarrow y e^{x}=(5-x) e^{-x}+e^{x}+c$

Divided by $e^{-x}$

\begin{aligned} &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c} \\ &\Rightarrow \mathrm{y}=-(\mathrm{x}-5)-1+\mathrm{ce}^{\mathrm{x}}-(1) \\ &\Rightarrow \mathrm{y}_{\mathrm{y}}=5-\mathrm{x}-1+\mathrm{ce}^{\mathrm{x}} \\ &\quad=4-\mathrm{x}+\mathrm{ce}^{\mathrm{x}} \end{aligned}

Since the curve passes through the point (0,2)

Put x=0 and

\begin{aligned} &y=2 \\ &2=6-0+c e^{\circ} \\ &\Rightarrow 2=6+c(1) \\ \end{aligned}                                  $[\therefore e^{0}=1]$

$\Rightarrow C=-4$

Put in (2) we get,
Equation of the curve is

$y=6-x-4 e^{x}$