Question

Show that the differential equations $(x-y)\frac{dy}{dx}=x+2y$ is homogeneous and solve it also.

OR

Find the differential equations of the family of curves $(x-h{)}^2+(y-k{)}^2=r^2,$ where h and k are arbitrary constants.

Answer 1

We have $(x-y)\frac{dy}{dx}=x+2y\Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}=f(x,y)say$

Put $x=\lambda x,y=\lambda y,f(\lambda x,\lambda y)=\frac{\lambda x+2\lambda y}{\lambda x-\lambda y}=\frac{x+2y}{x-y}=f(x,)$

Hence the differential equations is homogenous.

Now put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\therefore v+x\frac{dv}{dx}=\frac{x+2vx}{x-vx}=\frac{1+2v}{1-v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^2}{1-v}\Rightarrow \int \frac{1-v}{v^2+v+1}dv=\int \frac{dx}{x}.$

Let 1 – v =A $\frac{d}{dv}[v^2+v+1]+B\Rightarrow 1-v=A[2v+1]+B$

On equating the coefficients of like terms, we get : $A=-\frac{1}{2},B=\frac{3}{2}$

$\therefore \int (\frac{3}{2(v^2+v+1)}-\frac{1(2v+1)}{2(v^2+v+1)})dv=\int \frac{dx}{x}\Rightarrow \frac{3}{2}\int \frac{1}{(v+\frac{1}{2})+{\left(\frac{\sqrt{3}}{2}\right)}^2}dv-\frac{1}{2}log$
$\left|\begin{array}{c}{v}^2+v+1\end{array}\right|=log\left|\begin{array}{c}x\end{array}\right|$

$\Rightarrow \frac{3}{2}\times {\frac{2}{\sqrt{3}}}^{tan-1}\left(\frac{2v+1}{\sqrt{3}}\right)-\frac{1}{2}log\left|\begin{array}{c}\frac{y^2+xy+x^2}{x^2}\end{array}\right|=log\left|\begin{array}{c}x\end{array}\right|+C$

$\therefore {\sqrt{3}}^{tan-1\left(\frac{2y+x}{\sqrt{3}x}\right)}-\frac{1}{2}log\left|\begin{array}{c}{y}^2+xy+x^2\end{array}\right|=C$

OR

We have $(x-h{)}^2+(y-k{)}^2=r^2\dots \left(i\right)\Rightarrow 2(x-h)(1-0)+2(y-k)(y-0)=0$

$\Rightarrow (x-h)=-(y-k)y^{\prime }\dots \left(ii\right)\Rightarrow -1=(y-k)y"+(y^{\prime }{)}^2$

$\Rightarrow -\frac{[1+(y^{\prime }{)}^2]}{y""}=(y-k)$ Replacing value of (y – k) in (ii), (x – h) = $\frac{[1+(y^{\prime }{)}^2]}{y^{\prime }}\times y^{\prime }$

Substituing values of (x - h} and (y – k) in (i), we get :

$(\frac{[1+(y{)}^2]}{y^{\prime }}\times y^{\prime })+{(-\frac{[1+(y^{\prime }{)}^2]}{y^{\prime }})}^2=r^2$~~~~~$\Rightarrow \left(\right[1+\left(y^{\prime }{)}^2\right]y^{\prime 2})+[1+(y^{\prime }{)}^2{]}^2+=(r{y}^{\prime }{)}^2$

$\Rightarrow \left[\right(y^{\prime }{)}^2+1][1+(y^{\prime }{)}^2{]}^2=(ry"{)}^2\Rightarrow \frac{[1+(y^{\prime }{)}^2{]}^3}{y{"}^2}=r^3$

Hence $[1+y_1^2{]}^3=r^2{y}_2^2$ is the required differential equations of all circles of radius r .