Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 75 textbook solution.

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

Answers (1)

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}

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE introduces CCTV policy in board exams 2025; schools fixed as centres must have facility
anam-khan

CBSE gives parents, schools till October 16 to register students in Classes 9, 11
anam-khan

CBSE Board Exam 2025: Class 9, 11 registration begins today; fee details, guidelines

Latest Question