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Step-by-Step Solution
Step 1: Rewrite the Differential Equation
The given differential equation is
(t+1)\,dx = \bigl(2x + (t+1)^4\bigr)\,dt.
Divide both sides by (t+1) to get:
\frac{dx}{dt} = \frac{2x}{t+1} + (t+1)^3.
This is a first-order linear differential equation of the form
\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3.
Step 2: Identify the Integrating Factor
The integrating factor \mu(t) is given by
\mu(t) = e^{\int -\frac{2}{t+1}\,dt}.
Evaluate the integral in the exponent:
\int -\frac{2}{t+1}\,dt = -2 \ln\bigl(t+1\bigr).
Therefore,
\mu(t) = e^{-2\ln(t+1)} = (t+1)^{-2}.
Step 3: Multiply the Original Equation by the Integrating Factor
Multiply both sides of
\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3
by (t+1)^{-2} :
(t+1)^{-2}\,\frac{dx}{dt} - \frac{2}{t+1}(t+1)^{-2}\,x = (t+1)^{-2}(t+1)^3.
The left-hand side is the derivative of
(t+1)^{-2}x,
so we can rewrite the equation as
\frac{d}{dt}\Bigl((t+1)^{-2}x\Bigr) = (t+1).
Step 4: Integrate Both Sides
Integrate both sides with respect to t :
(t+1)^{-2} x = \int (t+1)\,dt = \frac{(t+1)^2}{2} + C,
where C is the constant of integration.
Step 5: Solve for x(t)
Multiply both sides by (t+1)^{2} :
x(t) = (t+1)^{2} \left(\frac{(t+1)^2}{2} + C\right)
= \frac{(t+1)^4}{2} + C(t+1)^2.
Step 6: Apply the Initial Condition
Given x(0) = 2 , substitute t = 0 into x(t) :
x(0) = \frac{(0+1)^4}{2} + C(0+1)^2
= \frac{1}{2} + C
= 2.
Solving for C :
C = 2 - \frac{1}{2} = \frac{3}{2}.
Step 7: Write the Final Function and Find x(1)
Substitute C = \frac{3}{2} back into x(t) :
x(t) = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2.
To find x(1) , set t = 1 :
x(1) = \frac{(1+1)^4}{2} + \frac{3}{2}(1+1)^2
= \frac{2^4}{2} + \frac{3}{2}\cdot 2^2
= 8 + 6
= 14.
Final Answer
x(1) = 14.
