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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 73 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 73 textbook solution.

Answers (1)

Answer:  -1+2e^{x^{2}/2}

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

Given that,

Slope of the tangent to the curve at point  (x , y) is equal to the sum of x coordinate and the product of x and y coordinate of that point

So our equation becomes,

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

Differential equation is of the form

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

Where

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

If

\begin{aligned} &=e^{P d \int x}\\ \\&=e^{-\int x d x}\\ \\&=e^{-x^{2} / 2} \end{aligned}

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}^{-\frac{\mathrm{x}^{2}}{2}}=\int \mathrm{xe}^{-\frac{\mathrm{x}^{2}}{2}} \mathrm{dx}+\mathrm{c} \end{aligned}

Putting ,

\begin{aligned} &-\frac{x^{2}}{2}=t \\ \\&\Rightarrow \frac{-2 x}{2} d x=d t \\ \\&\Rightarrow x d x=a b \end{aligned}

Thus our equation becomes

\begin{aligned} &y e^{-x^{2} / 2}=\int-e^{t} d t+c\\ &y e^{-x^{2} / 2}=-e^{t}+c\\ &\text { Putting back } t=-\frac{x^{2}}{2}\\ &\Rightarrow \mathrm{ye}^{-\frac{\mathrm{x}^{2}}{2}}\\ &=-\mathrm{e}^{-\mathrm{x} / 2}+\mathrm{c}\\ &\Rightarrow \frac{y}{\mathrm{e}^{\mathrm{x}^{2} / 2}} \end{aligned}

\begin{aligned} &=-\frac{1}{e^{x^{2} / 2}}+c \\ &\Rightarrow y=-1+c e^{x^{2 / 2}}-(1) \end{aligned}

Since curve passes through (0,1)

Putting

\begin{aligned} &\mathrm{x}=0, \\ \\&\mathrm{y}=1 \mathrm{in}(1) \\ \\&\mathrm{y}=-1+\mathrm{ce}^{\mathrm{x}^{2} / 2} \\ \\&1=-1+\mathrm{ce}^{\mathrm{a}_{2} / 2} \\ \\&\Rightarrow 1+1=\mathrm{C} \end{aligned}                                  [\because e^{0}=1]

\Rightarrow c=2

Putting value of c in (1)

\begin{aligned} &y=-1+c e^{x^{2 / 2}} \\ &y=-1+2 e^{x^{2 / 2}} \end{aligned}

\therefore Equation of the curve  is  -1+2e^{x^{2}/2}

Posted by

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