question_answer

Solve the following differential equation. \[(1+{{x}^{2}})\frac{dy}{dx}-2xy=({{x}^{2}}+2)({{x}^{2}}+1),\] When x = 1 and y = 2. If y is distance and x is the time, \[\frac{dy}{dx}\] is velocity. John rides a vehicle beyond the limit on a highway. What suggestion would you give him to help understands the risks of overlapping?

Answer:

Given differential equation is \[(1+{{x}^{2}})\frac{dy}{dx}-2xy=({{x}^{2}}+2)({{x}^{2}}+1)\] On dividing both sides by \[(1+{{x}^{2}}),\] we get \[\frac{dy}{dx}-\frac{2x}{1+{{x}^{2}}}y=\frac{({{x}^{2}}+2)({{x}^{2}}+1)}{(1+{{x}^{2}})}\] \[\Rightarrow \]   \[\frac{dy}{dx}-\frac{2x}{1+{{x}^{2}}}y={{x}^{2}}+2\] Which is a linear differential equation of the from             \[\frac{dy}{dx}+Py=Q\] Here, \[P=\frac{-\,2xy}{1+{{x}^{2}}}\] and \[Q={{x}^{2}}+2\] Then, integrating factor \[(IF)={{e}^{\int{P\,dx}}}={{e}^{\int{-\,\,\frac{2x}{1+{{x}^{2}}}}dx}}\] \[={{e}^{-\log \,|1\,\,+\,\,{{x}^{2}}|}}\]  \[\left[ \because \,\,\int{\frac{f'(x)}{f(x)}\,dx=\log \,f(x)} \right]\]             \[={{e}^{\log {{(|1\,\,+\,\,{{x}^{2}})}^{-1}}}}\]             \[=\frac{1}{1+{{x}^{2}}}\]                 \[[\because \,\,{{e}^{\log \,x}}=x]\]\[\therefore \] Solution of given differential equation is             \[y\cdot IF=\int{(Q.IF)\,dx+C}\]  \[\Rightarrow \] \[y\times \frac{1}{(1+{{x}^{2}})}=\int{({{x}^{2}}+2)}\frac{1}{(1+{{x}^{2}})}\,dx+C\] \[\Rightarrow \] \[\frac{y}{(1+{{x}^{2}})}=\int{\frac{{{x}^{2}}+1+1}{(1+{{x}^{2}})}}\,dx+C\] \[\Rightarrow \] \[\frac{y}{(1+{{x}^{2}})}=\int{\frac{1+{{x}^{2}}}{1+{{x}^{2}}}}\,dx+\int{\frac{1}{1+{{x}^{2}}}\,dx+C}\] \[\mathbf{Z=100x+120y}\] \[\frac{y}{1+{{x}^{2}}}=\int{1}\,dx+\int{\frac{1}{1+{{x}^{2}}}\,dx+C}\] \[\Rightarrow \] \[\frac{y}{1+{{x}^{2}}}=x+{{\tan }^{-1}}x+C\] \[\Rightarrow \]\[y=x(1+{{x}^{2}})+(1+{{x}^{2}}){{\tan }^{-1}}x+C(1+{{x}^{2}})\] ?(i) Also given that when x = 1, then y = 2. So, putting x = 1 and y = 2 in Eq. (i), we get             \[2=1(1+{{1}^{2}})+(1+{{1}^{2}})\,ta{{n}^{-1}}1+C(1+{{1}^{2}})\] \[\Rightarrow \]   \[2=2+2\left( \frac{\pi }{4} \right)\,+2\,C\] \[\Rightarrow \]   \[C=-\frac{\pi }{4}\] On putting the value of V in Eq. (i), we get \[y=x(1+{{x}^{2}})+(1+{{x}^{2}}){{\tan }^{-1}}x-\frac{\pi }{4}(1+{{x}^{2}})\] Which is the required solution. Value John would be told about the road safety and traffic regulation. Also, should be made understand that ?speed thrills but kills?.