Question

The general solution of the differential equation $x^2dy - 2xydx = x^4\cos\,x\, dx$ is

• A. $y = x^2 \sin\,x + cx^2$
• B. $y = x^2 \sin\,x + c$
• C. $y = \sin\,x + cx^2$
• D. $y = \cos \,x + cx^2$

Answer 1

Answer: (A)

$y = x^2 \sin\,x + cx^2$

Answer 2

$x^{2} d y-2 x y d x=x^{4} \cos\, x d x$ $\left(\frac{dy}{dx} = \frac{x^4 \ cos\, x+2xy}{x^2}\right)$ $\Rightarrow d y / d x-2 y / x=x^{2} \cos \,x$ I.F. $= e ^{\int-2 / x d x}=e^{-2 \log x}=1 / x^{2}$ Therefore, the general solution is $\left(y(\frac{1}{x^2}) = \int \frac{1}{x^2}(x^2 \ cos\, x)dx = sin\,x + c\right)$ $\therefore y = x ^{2}(\sin\, x + c )$ $=x^{2} \sin\, x+c x^{2}$