question_answer

Solve the given differential equation \[\sqrt{1-{{y}^{2}}}\,dx=({{\sin }^{-1}}y-x)\,dy,\] y(0) = 0.            

Answer:

We have, \[\sqrt{1-{{y}^{2}}}\,dx=({{\sin }^{-1}}y-x)\,dy\]             \[\Rightarrow \]   \[\frac{dx}{dy}=\frac{{{\sin }^{-1}}y}{\sqrt{1-{{y}^{2}}}}-\frac{x}{\sqrt{1-{{y}^{2}}}}\]             \[\Rightarrow \]   \[\frac{dx}{dy}+\frac{x}{\sqrt{1-{{y}^{2}}}}=\frac{{{\sin }^{-1}}y}{\sqrt{1-{{y}^{2}}}},\] Which is the linear equation of the from \[\frac{dx}{dy}+Px=Q.\] Here,     \[P=\frac{1}{\sqrt{1-{{y}^{2}}}},\] \[Q=\frac{{{\sin }^{-1}}y}{\sqrt{1-{{y}^{2}}}}\] \[\therefore \]      \[IF={{e}^{\int{pdy}}}{{e}^{\int{\frac{1}{\sqrt{1-{{y}^{2}}}}\,dy}}}={{e}^{{{\sin }^{-1}}y}}\] \[\therefore \] Solution of the differential equation is given by             \[x\cdot IF=\int{Q\cdot }IF\,dy+C\] \[\Rightarrow \]   \[x{{e}^{{{\sin }^{-1}}y}}=\int{\frac{{{\sin }^{-1}}y}{\sqrt{1-{{y}^{2}}}}{{e}^{{{\sin }^{-1}}y}}\,dy+C}\] \[\Rightarrow \]   \[x{{e}^{{{\sin }^{-1}}y}}=\int{\underset{I}{\mathop{t}}\,\,\underset{II}{\mathop{{{e}^{t}}}}\,}\,dt+C\]    \[\left[ \because \,\,\text{put}\,\,{{\sin }^{-1}}y=t\,\,\,\Rightarrow \,\,\,\frac{1}{\sqrt{1-{{y}^{2}}}}\,dy=dt \right]\] \[\Rightarrow \]   \[x{{e}^{{{\sin }^{-1}}y}}=t\int{{{e}^{t}}\,dt-\int{1}}\,{{e}^{t}}\,dt+C\]                                     [Integration by parts] \[\Rightarrow \]   \[x{{e}^{{{\sin }^{-1}}y}}={{e}^{t}}(t\,-1)+C\] \[\Rightarrow \]   \[x{{e}^{{{\sin }^{-1}}y}}={{e}^{{{\sin }^{-1}}y}}({{\sin }^{-1}}y-1)+C\] It is given that y (0) = 0, i.e. y = 0 when x = 0. \[\therefore \]      \[0={{e}^{0}}(0-1)+C\] \[\Rightarrow \] C = 1 \[\therefore \]      \[x{{e}^{{{\sin }^{-1}}y}}={{e}^{{{\sin }^{-1}}y}}({{\sin }^{-1}}y-1)+1\] \[\Rightarrow \]   \[{{e}^{{{\sin }^{-1}}y}}(x-{{\sin }^{-1}}y+1)=1\] Which is the required solution.