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Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 71 textbook solution.

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Answer: y+3=(x+4)^{2}


At any point (x , y)of a curve , the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3)


Using variable separable method and substituting the values.


Slope of the tangent to the curve  =\frac{dy}{dx}

Slope of line segment joining (x , y) and (-4,-3)

\begin{aligned} &=\frac{y_{2}-y_{1}}{-4+x} \\ \end{aligned}

\begin{aligned} &=\frac{-3-y}{-4-x} \end{aligned}

\begin{aligned} &=\frac{-(y+3)}{-(x+4} \\ \end{aligned}

\begin{aligned} &=\frac{y+3}{x+4} \end{aligned}

Given at point (x , y) . Slope of tangent is twice of line segment

\begin{aligned} &\frac{d y}{d x}=2\left(\frac{y+3}{x+4}\right)^{2} \\ &\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4} \end{aligned}

Integrating both sides.

\begin{aligned} &\int \frac{d y}{y+3}=2 \int \frac{d x}{x+4} \\ \end{aligned}

\begin{aligned} &\Rightarrow \log |y+3|=2 \log |x+4|+\log c \\ \end{aligned}

\begin{aligned} &\Rightarrow \log |y+3|=\log |x+4|^{2}+\log c\left[\therefore a \log x=\log x^{2}\right] \\ \end{aligned}

\begin{aligned} &\Rightarrow \log (y+3)-\log (x+4)^{2}=\log c \\ \end{aligned}

\begin{aligned} &\log \frac{(y+3)}{(x+4)^{2}}=\log c \\ \end{aligned}

\begin{aligned} &\Rightarrow \frac{y+3}{(x+4)^{2}}=c-(1) \end{aligned}

Since curve passes through (-2,1)
Put x=-2 \; and\; y=1 in (1)

\begin{gathered} \frac{1+3}{(-2+4)^{2}}=C \\ \end{gathered}

\begin{gathered} \Rightarrow C=\frac{4}{(2)^{2}} \\ \end{gathered}

      \begin{gathered} =\frac{4}{4} \\ =1 \\ \Rightarrow C=1 \end{gathered}

Put value of c in eq (1)

\begin{aligned} &\frac{y+3}{(x+4)^{2}}=1 \\ &\Rightarrow y+3=(x+4)^{2} \end{aligned}

Hence the equation of the curve is


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