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

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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


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