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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 34

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 34

Answers (1)

Answer:x+y \log y=y

Given: The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact.

To find: The particular curve through the point \left ( 1,1 \right )

Hint: Use equation of tangent i.e. Y-y=\frac{d y}{d x}(X-x)  and compare with  \frac{d x}{d y}+P y=Q

Solution: Let P(x,y)  be the point in the curve y=y(x) such that the tangent at P cuts the co-ordinate at A and B. We know that the equation of tangent Y-t=\frac{d y}{d x}(X-x)

Putting Y=0 then

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

Co-ordinate of  B=\left(-y \frac{d x}{d y}+x, 0\right)

Here, x intercept of tangent = y

        =-y \frac{d x}{d y}+x=y

        =\frac{d x}{d y}-\frac{x}{y}-1

This is a linear differential equation

Now comparing with \frac{d x}{d y}+P y=Q

        P=\frac{1}{y}, Q=-1

        \begin{aligned} &=I f=e^{\int P d y} \\\\ &=I f=e^{\int \frac{1}{y} d y} \end{aligned}

        \begin{aligned} &=I f=e^{\log y} \\\\ &=I f=\frac{1}{y} \end{aligned}

So solution of equation is given by

        \begin{aligned} &=x(I f)=\int \mathrm{Q}(I f) d y+C \\\\ &=x\left(\frac{1}{y}\right)=\int(-1)\left(\frac{1}{y}\right) d y+C \\\\ &=\frac{x}{y}=-\log y+C \ldots(i) \end{aligned}

As it passes through the point \left ( 1,1 \right )

        \begin{aligned} &=\frac{1}{1}=-\log 1+C \\\\ &=1=C \quad[\log 1=0] \end{aligned}

Substituting value of C in equation (i) we get

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

Hence, required equation of curve is found.

 

 

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