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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 76 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 76 textbook solution.

Answers (1)

Answer : y^{2}=x+5

Given:  The slope of the tangent to the curve at any point is the reciprocal of twice the ordinate at that point. Also the curve passes through the point(4,3)

Hint: Using variable separable method

Explanation: Let P (x , y) be any point on the curve

Slop of the tangent at P (x , y)=\frac{dy}{dx}

\therefore Acc to given condition,

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \\ \\&\Rightarrow 2 \mathrm{ydy}=1 \mathrm{dx} \end{aligned}

Integrating both sides,

\begin{aligned} &2 \mathrm{ydy}=\int 1 \mathrm{dx} \\ \\&\Rightarrow \frac{2 \mathrm{y}^{2}}{2}=\mathrm{x}+\mathrm{c} \\ \\&\Rightarrow \mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \end{aligned}

Since the curve passes through (4,3) then,

\\a=4+c\\\Rightarrow C=5

Hence the required equation of the curve is

y^{2}=x+5

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