Find the general solution of the differential equation cosec x $\frac{d^{2}y}{dx^{2}} = x$ .

1. Rearrange the differential equation to express the second derivative explicitly.

$\frac{d^2y}{dx^2} = \frac{x}{\text{cosec } x} = x \sin x$

2. Integrate once with respect to $x$ using integration by parts to find $\frac{dy}{dx}$.

$\frac{dy}{dx} = \int x \sin x \, dx$
Using integration by parts, let $u = x$ and $\frac{dv}{dx} = \sin x$.
Then $\frac{du}{dx} = 1$ and $v = -\cos x$.
$\frac{dy}{dx} = -x \cos x - \int (-\cos x) \, dx$
$\frac{dy}{dx} = -x \cos x + \sin x + A$

3. Integrate a second time using integration by parts to find the general solution for $y$.

$y = \int (-x \cos x + \sin x + A) \, dx$
Evaluate $\int x \cos x \, dx$ using integration by parts:
Let $u = x$ and $\frac{dv}{dx} = \cos x$, giving $\frac{du}{dx} = 1$ and $v = \sin x$.
$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x$
Substitute this back into the integral for $y$:
$y = -(x \sin x + \cos x) + (-\cos x) + Ax + B$
$y = -x \sin x - 2\cos x + Ax + B$

1. Find the first derivative by integrating the rearranged differential equation using integration by parts.

Rearranging the differential equation from part (a):
$\frac{d^{2}y}{dx^{2}} = x \sin x$
Integrating with respect to $x$ using integration by parts:
$\frac{dy}{dx} = \int x \sin x \, dx$
Let $u = x$ and $dv = \sin x \, dx$, then $du = dx$ and $v = -\cos x$.
$\frac{dy}{dx} = -x \cos x - \int (-\cos x) \, dx$
$\frac{dy}{dx} = -x \cos x + \sin x + C$

2. Integrate the first derivative to find the general equation for $y$.

Integrating again to find $y$:
$y = \int (-x \cos x + \sin x + C) \, dx$
For $\int x \cos x \, dx$, use integration by parts again ($u = x, dv = \cos x \, dx \implies du = dx, v = \sin x$):
$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x$
Therefore:
$y = - (x \sin x + \cos x) + \int \sin x \, dx + \int C \, dx$
$y = -x \sin x - \cos x - \cos x + Cx + D$
$y = -x \sin x - 2\cos x + Cx + D$

3. Apply the geometric condition for the tangent to find the integration constant $C$.

The tangent at the origin $(0, 0)$ is parallel to the line $y = 3x - 5$, meaning the gradient at $x = 0$ is $3$.
Substitute $x = 0$ and $\frac{dy}{dx} = 3$ into the first derivative equation:
$3 = -0 \cos 0 + \sin 0 + C$
$C = 3$

4. Substitute the coordinates of the origin to find the constant $D$ and state the final equation.

Since the curve passes through the origin $(0,0)$, substitute $x = 0$, $y = 0$, and $C = 3$ into the general equation:
$0 = -0 \sin 0 - 2\cos 0 + 3(0) + D$
$0 = -2(1) + D \implies D = 2$
The equation of the solution curve is:
$y = -x \sin x - 2\cos x + 3x + 2$