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  • 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}

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