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 \(x(1)\):

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

Answer 2

Find \( x(1) \) given the differential equation \( (t + 1) dx = \left(2x + (t + 1)^4\right) dt \), with initial condition \( x(0) = 2 \).

Concept Used:

This is a first order linear differential equation. The standard form is \( \frac{dx}{dt} + P(t)x = Q(t) \). The solution is found using the integrating factor method, where the integrating factor (I.F.) is \( e^{\int P(t) dt} \). The general solution is then given by \( x \cdot \text{(I.F.)} = \int Q(t) \cdot \text{(I.F.)} dt + C \).

Step-by-Step Solution:

Step 1: Rewrite the given differential equation in the standard linear form.

\[ (t + 1) dx = \left(2x + (t + 1)^4\right) dt \] \[ \frac{dx}{dt} = \frac{2x + (t + 1)^4}{t + 1} \] \[ \frac{dx}{dt} = \frac{2}{t+1}x + (t+1)^3 \] \[ \frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3 \]

This is now in the form \( \frac{dx}{dt} + P(t)x = Q(t) \), where \( P(t) = -\frac{2}{t+1} \) and \( Q(t) = (t+1)^3 \).

Step 2: Find the Integrating Factor (I.F.).

\[ \text{I.F.} = e^{\int P(t) dt} = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln|t+1|} = e^{\ln((t+1)^{-2})} = (t+1)^{-2} \]

Step 3: Multiply the standard form equation by the Integrating Factor.

\[ (t+1)^{-2} \frac{dx}{dt} - \frac{2}{(t+1)^3}x = (t+1)^3 \cdot (t+1)^{-2} \] \[ \frac{d}{dt}\left[ x \cdot (t+1)^{-2} \right] = (t+1) \]

The left side is the derivative of \( x \cdot \text{(I.F.)} \).

Step 4: Integrate both sides with respect to \( t \).

\[ \int \frac{d}{dt}\left[ x \cdot (t+1)^{-2} \right] dt = \int (t+1) dt \] \[ x \cdot (t+1)^{-2} = \frac{(t+1)^2}{2} + C \]

Where \( C \) is the constant of integration.

Step 5: Apply the initial condition \( x(0) = 2 \) to find the constant \( C \).

\[ 2 \cdot (0+1)^{-2} = \frac{(0+1)^2}{2} + C \] \[ 2 \cdot 1 = \frac{1}{2} + C \] \[ 2 - \frac{1}{2} = C \implies C = \frac{3}{2} \]

Step 6: Write the particular solution of the differential equation.

\[ x \cdot (t+1)^{-2} = \frac{(t+1)^2}{2} + \frac{3}{2} \] \[ x = (t+1)^{2} \left( \frac{(t+1)^2}{2} + \frac{3}{2} \right) \] \[ x = \frac{(t+1)^4 + 3(t+1)^2}{2} \]

Step 7: Substitute \( t = 1 \) to find \( x(1) \).

\[ x(1) = \frac{(1+1)^4 + 3(1+1)^2}{2} \] \[ x(1) = \frac{(2)^4 + 3(2)^2}{2} \] \[ x(1) = \frac{16 + 3 \cdot 4}{2} = \frac{16 + 12}{2} = \frac{28}{2} = 14 \]

Hence, the value of \( x(1) \) is 14.