The solution of the differential equation $ydx - xdy + \log x dx = 0$ is

Let $y=y(x)$ be the solution of the differential equation $(x^2+1) y^{\prime}-2 x y=(x^4+2 x^2+1) \cos x$,with $y(0)=1$. Then $\int_{-3}^3 y(x) d x$ is :

$A$ function $y=f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with the condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to

The integrating factor of the differential equation $x \frac{dy}{dx} - y = x^2$ is . . . . . . .

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the tangent line $Y-y=Y^{\prime}(x)(X-x)$ and the coordinate axes,where $(x, y)$ is any point on the curve,is always $\frac{-y^2}{2 Y^{\prime}(x)}+1$,where $Y^{\prime}(x) \neq 0$. If $Y(1)=1$,then $12 Y(2)$ equals

Let the slope of the tangent line to a curve at any point $P(x, y)$ be given by $\frac{xy^2 + y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$,then the value of $y$,for which the point $(3, y)$ lies on the curve,is ..... .