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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 15

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 15

Answers (1)

Answer: \tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)

Given: The angle is \left(\frac{y}{x}-\frac{y}{x}\right)

To find: We have to find the equation of the curve which passes through \left(1, \frac{\pi}{4}\right)

Hint: First take the slope of the curve i.e.\frac{d y}{d x}=\tan \theta and take a linear equation y=v x

Solution:

The slope of the curve is \frac{d y}{d x}=\tan \theta

We have,

        =\theta=\left[\frac{y}{x}-\frac{y}{x}\right]

        \begin{aligned} &\text { Then } \begin{array}{l} \frac{d y}{d x}=\text { tan inverse tan }\left\{\left[\frac{y}{x}-\frac{y}{x}\right]\right\} \\\\ \quad=\frac{d y}{d x}=\frac{y}{x}-\frac{y}{x} \end{array} \\ &\text { Let } y=v x \end{aligned}

Differentiating with respect to x we get

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

        x \frac{d v}{d x}=\frac{d y}{d x}-\frac{y}{x} \quad\quad\quad\quad\left[v=\frac{y}{x}\right]

        x \frac{d v}{d x}=-v        [From equation i]

Integrating on both sides

        \begin{aligned} &\int v d v=\int-\frac{1}{x} d x \\\\ &\tan \frac{y}{x}=-\log |x|+C \ldots(i i) \end{aligned}

The curve passes through\left[1, \frac{\pi}{4}\right] it satisfices equation (ii)

        \begin{aligned} &=\log \frac{\pi}{4}=-\log |1|+C \\\\ &=C=1 \end{aligned}

\begin{aligned} &\text { Put } C=1, \tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=\log \left|\frac{e}{x}\right| \end{aligned}

Hence, required equation is found.

 

 

Posted by

infoexpert26

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