Question

If $x = x(t)$ is the solution of the differential equation $(t + 1) dx = \left(2x + (t + 1)^4\right) dt, \quad x(0) = 2,$ then $x(1)$ equals ____.

Answer 1

We are given the differential equation:

$(t + 1) \frac{dx}{dt} = 2x + (t + 1)^4$

To solve this, divide both sides by $(t + 1)$:

$\frac{dx}{dt} = \frac{2x}{t + 1} + (t + 1)^3$

Now, separate the variables:

$\frac{dx}{2x} = \frac{1}{t + 1} dt + (t + 1)^2 dt$

We can now integrate both sides:

$\int \frac{1}{2x} dx = \int \left(\frac{1}{t + 1} + (t + 1)^2\right) dt$

The left-hand side gives:

$\frac{1}{2} \ln |x|$

For the right-hand side, integrate each term:

$\int \frac{1}{t + 1} dt = \ln |t + 1| \quad \text{and} \quad \int (t + 1)^2 dt = \frac{(t + 1)^3}{3}$

Thus, we have:

$\frac{1}{2} \ln |x| = \ln |t + 1| + \frac{(t + 1)^3}{3} + C$

Exponentiate both sides:

$|x| = e^{2 \ln |t + 1| + \frac{2(t + 1)^3}{3} + 2C}$

Simplify:

$x = A(t + 1)^2 e^{\frac{2(t + 1)^3}{3}}$

Now, use the initial condition $x(0) = 2$:

$x(0) = A(1)^2 e^0 = 2 \implies A = 2$

Thus, the solution is:

$x = 2(t + 1)^2 e^{\frac{2(t + 1)^3}{3}}$

Finally, calculate $x(1)$:

$x(1) = 2(1 + 1)^2 e^{\frac{2(2)^3}{3}} = 2(2)^2 e^{\frac{16}{3}} = 2 \times 4 \times e^{\frac{16}{3}} \approx 14$