The solution of $\frac{d^2y}{dx^2} = \cos x - \sin x$ is

Which one of the following curves represents the solution of the initial value problem $Dy = 100 - y$,where $y(0) = 50$?

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$, where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

Verify that the given function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$.

$A$ function $y = f(x)$ satisfies the condition $f'(x) \sin x + f(x) \cos x = 1$,where $f(x)$ is bounded as $x \rightarrow 0$. If $I = \int_{0}^{\frac{\pi}{2}} f(x) \, dx$,then:

$A$ solution of $y = 2x\left( \frac{dy}{dx} \right) + x^2\left( \frac{dy}{dx} \right)^4$ is