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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

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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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