question_answer

Show that the differential equation \[\left[ x\,{{\sin }^{2}}\left( \frac{y}{x} \right)-y \right]\,dx+x\,dy=0\]

Is homogeneous. Find the particular solution of this differential equation, given that \[y=\frac{\pi }{4},\] when x = 1.

OR

Find the solution of differential equation

\[{{x}^{2}}dy+y(x+y)dx=0,\] if x = 1 and y = 1.

Answer:

Given differential equation is

\[\left[ x{{\sin }^{2}}\left( \frac{y}{x} \right)-y \right]dx+x\,dy=0\]

\[\Rightarrow \]   \[\frac{dy}{dx}=\frac{y-x{{\sin }^{2}}\left( \frac{y}{x} \right)}{x}\]                      ?(i)

Let \[F\,(x,\,\,y)=\frac{y-x{{\sin }^{2}}\left( \frac{y}{x} \right)}{x}\]

Then, \[F\,(\lambda x,\,\,\lambda y)=\frac{\lambda y-\lambda x{{\sin }^{2}}\left( \frac{\lambda y}{\lambda x} \right)}{\lambda x}\]

\[=\frac{\lambda \left[ y-x{{\sin }^{2}}\left( \frac{y}{x} \right) \right]}{\lambda x}={{\lambda }^{0}}F\,(x,\,\,y)\]

So, \[\frac{dy}{dx}=F\,(x,\,\,y)\]is a homogeneous differential equation.

Put \[y=vx\]and \[\frac{dy}{dx}=v+x\frac{dv}{dx}=v+x\frac{dv}{dx}\]in Eq. (i), we get

\[v+x\frac{dv}{dx}=\frac{vx-x{{\sin }^{2}}\left( \frac{vx}{x} \right)}{x}\]

\[\Rightarrow \]\[v+x\frac{dv}{dx}=v-{{\sin }^{2}}\,\,(v)\Rightarrow x\frac{dv}{dx}=-{{\sin }^{2}}\,\,(v)\]

\[\Rightarrow \]\[\cos e{{c}^{2}}\,\,v\,dv=-\frac{dx}{x}\]

On integrating both sides, we get

\[\int{\cos e{{c}^{2}}\,v}+\int{\frac{dx}{x}=0}\]

\[\Rightarrow \]   \[-\cot \,\,v+\log |x|=C\]

\[\Rightarrow \]\[-\cot \left( \frac{y}{x} \right)+\log |x|=C\]            \[\left[ put\,\,v=\frac{y}{x} \right]\]          ?(ii)

Also given that\[y=\frac{\pi }{4},\] when \[x=1\]

\[\therefore \]\[-\cot \,\,\frac{\pi }{4}+\log |1|=C\Rightarrow C=-\,1+0\Rightarrow C=-\,1\]

So, the required particular solution is\[-\,\cot \left( \frac{y}{x} \right)+\log |x|\,\,=-\,1\]

\[\Rightarrow \]   \[1+\log |x|-\cot \left( \frac{y}{x} \right)=0\].

Or

Given differential equation\[{{x}^{2}}dy+y\,\,(x+y)\,\,dx=0\]can be written as\[{{x}^{2}}dy+(xy+{{y}^{2}})\,\,dx=0\]

\[\Rightarrow \]   \[{{x}^{2}}dy=-(xy+{{y}^{2}})\,\,dx\]

\[\Rightarrow \]   \[\frac{dy}{dx}=-\left( \frac{yx+{{y}^{2}}}{{{x}^{2}}} \right)\]

\[\Rightarrow \]   \[\frac{dy}{dx}=-\left( \frac{y}{x} \right)-{{\left( \frac{y}{x} \right)}^{2}}\]                         ?(i)

which is a homogeneous, as\[\frac{dy}{dx}=f\,\,\left( \frac{y}{x} \right)\].

On putting \[y=vx\]and \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]in Eq. (i), we get

\[v+x\frac{dv}{dx}=-\,v-{{v}^{2}}\]

\[\Rightarrow \]   \[x\frac{dv}{dx}=-\,2v-{{v}^{2}}\]

\[\Rightarrow \]   \[\frac{1}{2v+{{v}^{2}}}dv=-\frac{1}{x}dx\]

\[\Rightarrow \]   \[\frac{1}{v\,\,(2+v)}dv=-\frac{1}{x}dx\]

\[\Rightarrow \]   \[\frac{2}{2v\,\,(2+v)}dv=-\frac{1}{x}dx\]

\[\Rightarrow \]\[\frac{1}{2}\,\,\left( \frac{1}{v}-\frac{1}{v+2} \right)\,\,dv=-\frac{1}{x}dx\]

\[\Rightarrow \]\[\frac{1}{2}\,\,\int{\,\,\frac{1}{v}\,\,dv}-\frac{1}{2}\,\,\int{\,\,\frac{1}{v+2}\,\,}dv=-\int{\,\,\frac{1}{x}\,\,dx}+\log C\]

\[\Rightarrow \]\[\frac{1}{2}\log |v|-\frac{1}{2}\log |v+2|\,\,=-\,|\log x|+\log C\]

\[\Rightarrow \]   \[\frac{1}{2}\log \,\,\left| \frac{v}{v+2} \right|=\log \,\,\left| \frac{C}{x} \right|\]

\[\Rightarrow \]   \[\log \,\,\left| \frac{v}{v+2} \right|=2\,\log \,\,\left| \frac{C}{x} \right|\Rightarrow \frac{v}{v+2}={{\left( \frac{C}{x} \right)}^{2}}\]

\[\Rightarrow \]               \[\frac{\frac{y}{x}}{\frac{y}{x}+2}={{\left( \frac{C}{x} \right)}^{2}}\]\[\left[ put\,\,v=\frac{y}{x} \right]\]

\[\Rightarrow \]               \[\frac{y}{y+2x}={{\left( \frac{C}{x} \right)}^{2}}\]

It is given that y = 1, when x = 1.

\[\therefore \]                  \[\frac{1}{1+2}={{\left( \frac{C}{1} \right)}^{2}}\Rightarrow \frac{1}{3}={{C}^{2}}\]

Hence, the particular solution of the given differential equation is

\[\frac{y}{y+2x}=\frac{\frac{1}{3}}{{{x}^{2}}}\]

\[\Rightarrow \]               \[\frac{y}{y+2x}=\frac{1}{3{{x}^{2}}}\]

\[\Rightarrow \]   \[3{{x}^{2}}y=y+2x\]