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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 24 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 24 maths

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Answer: x^{2}+y^{2}=C x

Given: A curve is such that the length of the perpendicular from the origin in the tangent at any point P of the curve is equal to the abscissa of P.

To find: The differential equation of the curve y^{2}-2 x y \frac{d y}{d x}-x^{2}=0  also we we have to find the curve

Hint: if the differential equation is homogeneous then put  y=v x=\frac{d y}{d x}=v+x \frac{d v}{d x}

Solution: we have   y^{2}-2 x y \frac{d y}{d x}-x^{2}=0

        =\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}

It is a homogeneous differential equation

\begin{aligned} &\text { Put } y=v x \\ &\qquad=\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned}

\text { Now, } x \frac{d v}{d x}+v=\frac{v^{2} x^{2}-x^{2}}{2 x v x}

        =x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v        [Taking x common on right hand side]

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

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

Integrating on both sides we get

        \begin{aligned} &=\int \frac{2 v d v}{v^{2}+1}=-\int \frac{d x}{x} \\\\ &=\int \frac{2 v d v}{v^{2}+1}=-\int \frac{d x}{x} \quad\left[t=v^{2}+1, d t=2 v d v\right] \\\\ &=\int \frac{d t}{t}=-\int \frac{d x}{x} \end{aligned}

        \begin{aligned} &=\log t=-\log x+\log C \\\\ &=\log \left(v^{2}+1\right)=-\log x+\log C \end{aligned}

        =\log \left(v^{2}+1\right)=\log \frac{c}{x} \quad=v^{2}+1=\frac{c}{x}

        =\left(\frac{y}{x}\right)^{2}+1=\frac{c}{x} \quad\quad\quad\quad\left[y=v x, v=\frac{y}{x}\right]

        \begin{aligned} &=\frac{y^{2}}{x^{2}}+1=\frac{C}{x} \\\\ &=\frac{y^{2}+x^{2}}{x^{2}}=\frac{C}{x} \end{aligned}

        \begin{aligned} &=y^{2}+x^{2}=\frac{C x^{2}}{x} \\\\ &=y^{2}+x^{2}=C x \end{aligned}

Differentiating with respect to x

        =2 x+2 y \frac{d y}{d x}-C=0

        =\frac{d y}{d x}=\frac{C-2 x}{2 y}

Let \left ( h,k \right ) be the point where tangent passes through origin and length is equal to h. So, equation of tangent at (h, k) is

        \begin{aligned} &=(y-k)=\left(\frac{d y}{d x}\right)_{(h, k)}(x-h) \\\\ &=y-k=\left(\frac{c-2 h}{2 k}\right)(x-h) \end{aligned}

        \begin{aligned} &=2 k y-2 k^{2}=x C-2 h x-h C+2 h^{2} \\\\ &=x(C-2 h)-2 k y+2 k^{2}-h C+2 h^{2}=0 \end{aligned}

        =x(C-2 h)-2 k y+2\left(k^{2}+h^{2}\right)-h C=0

        =x(C-2 h)-2 k y+2(C h)-h C=0 \quad\quad\quad\quad\left[h^{2}+k^{2}=C h \text { on the curve }\right]

        =x(C-2 h)-2 k y++C h=0

Length of perpendicular as tangent from origin is

        =L=\left[\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right]

        \begin{aligned} &=\left[\frac{(0)(C-2 h)+(0)(-2 k)+h C}{\sqrt{(C-2 h)^{2}+(-2 k)^{2}}}\right] \\\\ &=\frac{C h}{\sqrt{C^{2}+4 h^{2}+4 k^{2}-4 C h}} \end{aligned}

        \begin{aligned} &=\frac{C h}{\sqrt{C^{2}+4\left(h^{2}+k^{2}-C h\right)}} \\\\ &=\frac{C h}{\sqrt{C^{2}+4(0)}} \end{aligned}

        \begin{gathered} =\frac{C h}{\sqrt{C^{2}}} \\\\ =\frac{C h}{C} \\\\ L=C \end{gathered}

Hence, x^{2}+y^{2}=C x is the required curve.

 

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