Question

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

• A. $32$
• B. $\frac{64}{3}$
• C. $\frac{128}{3}$
• D. $64$

Answer 1

The correct option is A $32$

We have

$xdy+2ydx=x\left(x-3\right)dxxdy=\left[x\left(x-3\right)-2y\right]dx\Rightarrow \frac{dy}{dx}=\frac{x\left(x-3\right)}{x}-\frac{2y}{x}\Rightarrow \frac{dy}{dx}+\frac{2y}{x}=\left(x-3\right)ItisanLDE\therefore If=e^{2\int \frac{1}{x}dx}=e^{2\log x}=x^{2}GeneralSolutiony\times x^2=\int \left(x-3\right)x^{2}dx+c{x}^{2}y=\int x^{3}dx-3\int x^{2}dx+c{x}^{2}y=\frac{x^4}{4}-x^3+cForthepoint\left(4,0\right)0=4-4+CC=0x^{2}y=\frac{x^4}{4}-x^{3}y=\frac{x^2}{4}-x$.