The particular solution of $e^{\frac{y}{x}} = x, y(1) = 3, x > 0$ is . . . . . . .

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

If $f: R \rightarrow R$ is a differentiable function such that $f^{\prime}(x) > 2f(x)$ for all $x \in R$,and $f(0) = 1$,then:

Let $x = x(y)$ be the solution of the differential equation $2(y + 2) \log_e(y + 2) dx + (x + 4 - 2 \log_e(y + 2)) dy = 0$,$y > -1$ with $x(e^4 - 2) = 1$. Then $x(e^9 - 2)$ is equal to

The solution of $\frac{dy}{dx} + y = e^x$ is

$A$ continuous function $f: R \rightarrow R$ satisfies the equation $f(x) = x + \int_0^x f(t) \, dt$. Which of the following options is true?