question_answer

A curve \[y=f(x)\] which passes through \[(4,0)\] satisfies the differential equation \[x\,dy+2y\,dx=x(x-3)\,dx.\]The area bounded by \[y=f(x)\] and line \[y=x\](in square unit) is

A) \[\frac{128}{3}\]                   

B)        \[\frac{64}{3}\]     

C) \[64\]                    

D)        \[128\]

Correct Answer: B

Solution :

   [b] \[x\,\,dy+2y\,\,dx=x\,(x-3)\,dx\] \[\Rightarrow \,\,\,\,\,{{x}^{2}}dy+2xy\,\,dx=({{x}^{3}}-3{{x}^{2}})dx\] \[\Rightarrow \,\,\,\,\,d({{x}^{2}}y)=({{x}^{3}}-3{{x}^{2}})dx\] \[\Rightarrow \,\,\,\,\,\,\,\,y{{x}^{2}}=\frac{{{x}^{4}}}{4}-{{x}^{3}}+c\] Curve passes through the point \[(4,0)\]. \[\therefore \,\,\,c=0\] Therefore, curve is\[y=\frac{{{x}^{2}}}{4}-x.\] Required area \[=\int\limits_{0}^{8}{\left( x-\left( \frac{{{x}^{2}}}{4}-x \right) \right)}dx\] \[=\left( {{x}^{2}}-\frac{{{x}^{3}}}{12} \right)_{0}^{8}=64-\frac{2}{3}\times 64=\frac{64}{3}\]