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

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

Answers (1)

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}

Posted by

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