Question

Consider the family of circles whose center lies on the straight line y = x.
If this family of circles is represented by the differential equation $Py''+Qy'+1=0$ , Where P, Q are functions of x, y and y’ (here $y'=\dfrac{dy}{dx},y''=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$), then which of the following is (are) true
$\begin{aligned} & \text{a) P = x + y} \\\ & \text{b) P = y - x} \\\ & \text{c) P + Q =1-x + y + y }\\!\\!'\\!\\!\text{ + (y }\\!\\!'\\!\\!\text{ }{{\text{)}}^{2}} \\\ & \text{d) P - Q = x + y - y }\\!\\!'\\!\\!\text{ - (y }\\!\\!'\\!\\!\text{ }{{\text{)}}^{2}} \\\ \end{aligned}$

Answer 1

We know that the center of all the circles lies on the line x = y. Let us take the centers to be (α, α) . Now we know that the equation of circle with centre (α, α) and radius r is ${{(x-\alpha )}^{2}}+{{(y-\alpha )}^{2}}={{r}^{2}}$ . Now we will differentiate the equation two times. From the first differential we find the value of α and substitute this in the second differential. Hence we will have an equation in the form of $Py''+Qy'+1=0$ from which we can find the value of P and Q.

Complete step by step answer:
Now we are given that the centers of the circles lie on the line x = y.
Hence let us take this centers as $(\alpha ,\alpha )$ .
Now we know that the equation of the circle with centre $(a,b)$ and radius $r$ is given by ${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$
Hence the equation of the circle with radius $(\alpha ,\alpha )$ and radius r will be.
${{(x-\alpha )}^{2}}+{{(y-\alpha )}^{2}}={{r}^{2}}$
Now Let us differentiate the equation with respect to x.
$2(x-\alpha )+2(y-\alpha )\dfrac{dy}{dx}=0$

& \Rightarrow (x-\alpha )+(y-\alpha )\dfrac{dy}{dx}=0..........(1) \\\ & \Rightarrow x-\alpha +y\dfrac{dy}{dx}-\alpha \dfrac{dy}{dx}=0 \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow x+y\dfrac{dy}{dx}-\alpha \left( 1+\dfrac{dy}{dx} \right)=0 \\\ & \Rightarrow x+y\dfrac{dy}{dx}=\alpha \left( 1+\dfrac{dy}{dx} \right) \\\ & \Rightarrow \dfrac{x+y\dfrac{dy}{dx}}{\left( 1+\dfrac{dy}{dx} \right)}=\alpha \\\ & \text{Hence we have }\alpha =\dfrac{x+yy'}{1+y'} \\\ & \alpha =\dfrac{x+yy'}{1+y'}.........................(2) \\\ \end{aligned}$$ Now let us again consider equation (1) which is $(x-\alpha )+(y-\alpha )\dfrac{dy}{dx}=0.$ Differentiating the equation again we get. $$(1)+\left[ {{\left( \dfrac{dy}{dx} \right)}^{2}}+\left( y-\alpha \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right]=0$$ Now substituting the value of α from (2) we get $$\begin{aligned} & (1)+\left[ y{{'}^{2}}+\left( y-\dfrac{x+yy'}{1+y'} \right)y'' \right]=0 \\\ & \Rightarrow 1+\left[ y{{'}^{2}}+\left( y-\dfrac{x+yy'}{1+y'} \right)y'' \right]=0 \\\ & \Rightarrow 1+y{{'}^{2}}+\left( \dfrac{y+yy'-x-yy'}{1+y'} \right)y''=0 \\\ & \Rightarrow 1+y{{'}^{2}}+\left( \dfrac{y-x}{1+y'} \right)y''=0 \\\ \end{aligned}$$ Multiplying the equation by $1+y'$ we get $\begin{aligned} & \Rightarrow 1+y{{'}^{2}}+y'+y{{'}^{3}}+(y-x)y''=0 \\\ & \Rightarrow y''(y-x)+y'(1+y'+y{{'}^{2}})+1=0 \\\ \end{aligned}$ Now Comparing the equation with $Py''+Qy'+1=0$ we get $\begin{aligned} & P=(y-x) \\\ & Q=(1+y'+y{{'}^{2}}) \\\ \end{aligned}$ $\begin{aligned} & P+Q=y-x+1+y'+y{{'}^{2}} \\\ & P-Q=y-x-1-y'-y{{'}^{2}} \\\ \end{aligned}$ **So, the correct answer is “Option B and C”.** **Note:** here while differentiating the equation of circle note that y is a function of x and hence differentiation of ${{(y-\alpha )}^{2}}=2(y-\alpha )\dfrac{dy}{dx}$ and not just $2(y-\alpha )$.