Math Problem Statement

Solve the initial value problem. t^2 dx/dt + 5tx = t^4 ln(t) + 4, x(1) = 0.

Solution

The initial value problem provided is:

t2dxdt+5tx=t4lnt+4,x(1)=0.t^2 \frac{dx}{dt} + 5tx = t^4 \ln t + 4, \quad x(1) = 0.

This is a first-order linear differential equation in x(t)x(t). To solve it, we'll follow the steps for solving linear differential equations:

Step 1: Write the equation in standard form

First, rewrite the equation:

dxdt+5xt=t2lnt+4t2.\frac{dx}{dt} + \frac{5x}{t} = t^2 \ln t + \frac{4}{t^2}.

Step 2: Find the integrating factor

The integrating factor μ(t)\mu(t) is found using the formula:

μ(t)=e5tdt=e5lnt=t5.\mu(t) = e^{\int \frac{5}{t} dt} = e^{5 \ln t} = t^5.

Step 3: Multiply the equation by the integrating factor

Multiply both sides of the equation by t5t^5:

t5dxdt+5t4x=t7lnt+4t3.t^5 \frac{dx}{dt} + 5t^4 x = t^7 \ln t + 4t^3.

The left-hand side is now the derivative of t5xt^5 x:

ddt(t5x)=t7lnt+4t3.\frac{d}{dt} (t^5 x) = t^7 \ln t + 4t^3.

Step 4: Integrate both sides

Integrate both sides with respect to tt:

t5x=(t7lnt+4t3)dt.t^5 x = \int (t^7 \ln t + 4t^3) dt.

We now compute the integrals:

t7lntdtand4t3dt.\int t^7 \ln t \, dt \quad \text{and} \quad \int 4t^3 \, dt.

  1. For t7lntdt\int t^7 \ln t \, dt, we use integration by parts, letting u=lntu = \ln t and dv=t7dtdv = t^7 dt. This gives: t7lntdt=t8lnt8t864.\int t^7 \ln t \, dt = \frac{t^8 \ln t}{8} - \frac{t^8}{64}.

  2. For 4t3dt\int 4t^3 \, dt, we simply get: 4t3dt=t4.\int 4t^3 \, dt = t^4.

Thus, the solution becomes:

t5x=t8lnt8t864+t4+C,t^5 x = \frac{t^8 \ln t}{8} - \frac{t^8}{64} + t^4 + C,

where CC is the constant of integration.

Step 5: Solve for x(t)x(t)

Now, solve for x(t)x(t) by dividing through by t5t^5:

x(t)=t3lnt8t364+1t+Ct5.x(t) = \frac{t^3 \ln t}{8} - \frac{t^3}{64} + \frac{1}{t} + \frac{C}{t^5}.

Step 6: Apply the initial condition

Use the initial condition x(1)=0x(1) = 0 to find CC:

0=13ln181364+11+C15.0 = \frac{1^3 \ln 1}{8} - \frac{1^3}{64} + \frac{1}{1} + \frac{C}{1^5}.

Since ln1=0\ln 1 = 0, this simplifies to:

0=164+1+C.0 = -\frac{1}{64} + 1 + C.

Thus,

C=6364.C = -\frac{63}{64}.

Final Solution:

Substitute C=6364C = -\frac{63}{64} into the expression for x(t)x(t):

x(t)=t3lnt8t364+1t6364t5.x(t) = \frac{t^3 \ln t}{8} - \frac{t^3}{64} + \frac{1}{t} - \frac{63}{64 t^5}.

This is the solution to the initial value problem.

Would you like further details on any step or have any questions?


Here are 5 related questions:

  1. What is an integrating factor, and why is it useful in solving linear differential equations?
  2. How does integration by parts work, and why was it needed for t7lntdt\int t^7 \ln t \, dt?
  3. Can you explain the significance of initial conditions in solving differential equations?
  4. How does the choice of the integrating factor simplify solving the equation?
  5. What other techniques can be used to solve first-order linear differential equations?

Tip: Always double-check initial conditions when solving differential equations to ensure the constants are correctly computed.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Integration by Parts
Logarithmic Functions

Formulas

dx/dt + (5x/t) = t^2 ln(t) + (4/t^2)
μ(t) = t^5
∫t^7 ln(t) dt
∫4t^3 dt

Theorems

Integrating Factor
Integration by Parts

Suitable Grade Level

University