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