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

Correct Answer: 14

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 \(x(1)\):

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