We have a first-order linear differential equation:

\[
t(t - 4) \frac{dy}{dt} + y = 0, \quad y(2) = 1
\]

To solve this differential equation, let's first rewrite it in standard linear form:

\[
\frac{dy}{dt} + \frac{1}{t(t - 4)} y = 0
\]

This is a linear differential equation of the form:

\[
\frac{dy}{dt} + p(t) y = 0,
\]

where \( p(t) = \frac{1}{t(t - 4)} \).

### Step 1: Find the Integrating Factor

The integrating factor \( \mu(t) \) is given by:

\[
\mu(t) = e^{\int p(t) \, dt} = e^{\int \frac{1}{t(t - 4)} \, dt}
\]

To compute this integral, we use partial fraction decomposition:

\[
\frac{1}{t(t - 4)} = \frac{A}{t} + \frac{B}{t - 4}
\]

Let's find \( A \) and \( B \) by equating:

\[
1 = A(t - 4) + Bt.
\]

Solving for \( A \) and \( B \), we get:

- For \( t = 4 \): \( 1 = 4B \Rightarrow B = \frac{1}{4} \)
- For \( t = 0 \): \( 1 = -4A \Rightarrow A = -\frac{1}{4} \)

Thus:

\[
\frac{1}{t(t - 4)} = -\frac{1}{4t} + \frac{1}{4(t - 4)}.
\]

Now, we integrate:

\[
\int \frac{1}{t(t - 4)} \, dt = \int \left( -\frac{1}{4t} + \frac{1}{4(t - 4)} \right) \, dt = -\frac{1}{4} \ln |t| + \frac{1}{4} \ln |t - 4| + C.
\]

Simplify the integrating factor:

\[
\mu(t) = e^{-\frac{1}{4} \ln |t| + \frac{1}{4} \ln |t - 4|} = \left( \frac{|t - 4|}{|t|} \right)^{1/4}.
\]

### Step 2: Solve the Differential Equation

Now, multiply the differential equation by the integrating factor \( \mu(t) \):

\[
\left( \frac{|t - 4|}{|t|} \right)^{1/4} \frac{dy}{dt} + \left( \frac{|t - 4|}{|t|} \right)^{1/4} \cdot \frac{1}{t(t - 4)} y = 0.
\]

Simplify to find \( y(t) \), and apply the initial condition \( y(2) = 1 \).

Would you like me to continue the solving process step-by-step?

Learn how to solve the first-order linear differential equation t(t-4)y’+y=0 with a detailed step-by-step solution. This problem involves differential equations, integrating factors, and partial fraction decomposition. Suitable for advanced undergraduate students.

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