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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

Answers (1)

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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