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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 16 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 16 maths

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Answer: \mathrm{c} y^{4}=e^{-x / y}

Given: The curve is y=f(x). SupposeP(x, y) is a part of the curve.

To find: We have to find the equation of the curve for which the intercept cut off by a tangent on x-axis.

Hint: Use equation of tangent i.e.Y-y=\frac{d y}{d x}(X-x) then solve the differential equation.

Solution: Equation of tangent to the curve at P is

=Y-y=\frac{d y}{d x}(X-x)

Where (X, Y) is the arbitrary point on the tangent

Putting Y=0 we get

\begin{aligned} &=0-y=\frac{d y}{d x}(X-x) \\\\ &=X-x=-y \frac{d x}{d y} \end{aligned}

When cut off by the tangent on the x-axis

=x-y \frac{d x}{d y}=4 y

Therefore   -y \frac{d x}{d y}=4 y-x

        \begin{aligned} &=\frac{d x}{d y}=\frac{x-4 y}{y} \\\\ &=\frac{d y}{d x}=\frac{y}{x-4 y} \ldots(i) \end{aligned}

This is homogeneous differential equation

Putting y=v x and \frac{d y}{d x}=v+x \frac{d v}{d x}  in (i)

We get

        \begin{aligned} &=v+x \frac{d v}{d x}=\frac{v x}{x-4 v x}=\frac{v x}{x(1-4 v)} \\\\ &=v+x \frac{d v}{d x}=\frac{v}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v}{1-4 v}-v \end{aligned}

        \begin{aligned} &=x \frac{d v}{d x}=\frac{v-v(1-4 v)}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v-v+4 v^{2}}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{4 v^{2}}{1-4 v} \end{aligned}

        \begin{aligned} &=\frac{1}{x} \frac{d x}{d v}=\frac{1-4 v}{4 v^{2}} \\\\ &=4 \frac{d x}{x}=\frac{1-4 v}{v^{2}} d v \end{aligned}

Integrating on both sides,

        \begin{aligned} &=4 \int \frac{d x}{x}=\int \frac{1-4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-\int \frac{4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-4 \int \frac{d v}{v^{2}} \end{aligned}

        \begin{aligned} &=4 \log x+\log C=-\frac{1}{v}-4 \log v \\\\ &=4 \log x+4 \log v+\log C=-\frac{1}{v} \\\\ &=4 \log (x v)+\log C=-\frac{1}{v} \end{aligned}

Substituting value of v we get

        \begin{aligned} &=4 \log \left(x \times \frac{y}{x}\right)+\log C=-\frac{x}{y} \\\\ &=4 \log y+\log C=-\frac{x}{y} \end{aligned}

        =\log y^{4}+\log C=-\frac{x}{y} \quad\quad\quad\quad\left[a \log x=\log x^{a}\right]

        \begin{aligned} &=\log \left(y^{4} C\right)=-\frac{x}{y} \\\\ &=y^{4} C=e^{-\frac{x}{y}} \end{aligned}

Hence, required curve is found.

Posted by

infoexpert26

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