If $y=y(x)$ is a particular solution of $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$ with $y(0)=1$,then $y\left(\frac{1}{2}\right) = $

Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$,and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $f(x)$ is

Suppose $y=y(x)$ is the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ such that $\lim_{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$- and $y$-intercepts of the tangent to the curve at $x=0$,then the value of $a-4b$ is equal to:

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$ under the condition $y = 1$ when $x = e$ is

An integrating factor for the differential equation $(1 + y^2)dx - (\tan^{-1} y - x)dy = 0$ is:

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is: