Solve the differential equation $$\frac{dr}{dx} = \frac{\ln x}{x^{2}}$$ for $x > 0$ given $x = 1$ when $t = 2$ Write your answer in the form $t^{2} = f(x)$. - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 2

Integrating the left side gives:

$r = \int \frac{\ln x}{x^2} dx$

For the right side, we use integration by parts where:

• Let $u = \ln x$ and $dv = \frac{dx}{x^2}$, so $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$.

Using integration by parts:

$r = -\frac{\ln x}{x} - \int -\frac{1}{x}\cdot \frac{1}{x} dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} dx$

Now integrating $\int \frac{1}{x^2} dx$ gives:

$\int \frac{1}{x^2} dx = -\frac{1}{x} + C$

Thus, we find:

$r = -\frac{\ln x}{x} - \frac{1}{x} + C$

Given $x = 1$ when $t = 2$, we substitute:

$2 = -\frac{\ln(1)}{1} - 1 + C$

This simplifies to:

$2 = 0 - 1 + C \Rightarrow C = 3$

Substituting $C$ back into our equation:

$r = -\frac{\ln x}{x} - \frac{1}{x} + 3$

Now, we need to express $t^2$ in terms of $x$:

Since $t = \sqrt{r}$, we have:

$t = \sqrt{-\frac{\ln x}{x} - \frac{1}{x} + 3}$

Thus:

$t^2 = -\frac{\ln x}{x} - \frac{1}{x} + 3$.