The solution of the differential equation $\frac{dy}{dx} = x^2 + x + \frac{1}{x}$ is

The solution of the differential equation $\frac{dy}{dx} + y = 1, y(1) = 0$ is

The solution of the initial value problem $x\frac{dy}{dx} = 2, y(1) = 2$ is

Which of the following are linear first order differential equations?

(A) $\frac{dy}{dx} + P(x)y = Q(x)$

(B) $\frac{dx}{dy} + P(y)x = Q(y)$

(C) $(x - y)\frac{dy}{dx} = x + 2y$

(D) $(1 + x^2)\frac{dy}{dx} + 2xy = 2$

Choose the correct answer from the options given below:

The general solution of the differintial equation : $x\frac{dy}{dx}-ay=1(a\neq0)$ is

Let $e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0$, where $C \in \mathbb{R}$ be a particular solution of the differential equation $x(e^{2y} - 1)dy + (x^2 - 1)e^ydx = 0$ and passes through the point $(1, 1)$. The value of $(\alpha + \beta + \gamma + \delta - C)$ is

A integrating factor of the differential equation $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x^2}$, $(x > 0)$ is equal to