Question

The solution of the differential equation \( \frac{dx}{dt} = x^2 \) with \( x(0) = 1 \) will tend to infinity as ____ .

• A. as \( t \to 1 \)
• B. as \( t \to 2 \)
• C. as \( t \to 0.5 \)
• D. as \( t \to \infty \)

Hint:

For differential equations of the form \( \frac{dx}{dt} = x^n \), solutions often diverge as \( t \to \infty \), especially when \( n \geq 1 \).

Answer 1

The Correct Option is D

To solve the differential equation \( \frac{dx}{dt} = x^2 \), we use the method of separation of variables. Rearranging the equation: \[ \frac{dx}{x^2} = dt \] Now, integrate both sides: \[ \int \frac{1}{x^2} dx = \int dt \] The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), and the integral of \( dt \) is \( t \). Thus, we have: \[ -\frac{1}{x} = t + C \] Using the initial condition \( x(0) = 1 \), we find \( C \): \[ -\frac{1}{1} = 0 + C \implies C = -1 \] Thus, the solution to the differential equation is: \[ -\frac{1}{x} = t - 1 \implies x = \frac{1}{1 - t} \] As \( t \to 1 \), the denominator approaches zero, causing \( x \) to tend to infinity. Therefore, the solution tends to infinity as **\( t \to \infty \)**.