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

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.