Give the solution 
y
(
t
)
y(t) of the initial-value problem 
t
d
y
d
t
+
6
y
=
3
t
−
2
t 
dt
dy
​
 +6y=3t 
−2
  with 
t
>
0
t>0 and 
y
(
1
)
=
0
y(1)=0.



y
(
t
)
=
y(t)=

We are given the first-order linear differential equation:

\[
t \frac{dy}{dt} + 6y = 3t - 2, \quad t > 0
\]
with the initial condition:

\[
y(1) = 0.
\]

### Step 1: Rewrite the equation in standard form
First, divide the entire equation by \(t\) to express it in the standard linear form:

\[
\frac{dy}{dt} + \frac{6}{t}y = 3 - \frac{2}{t}.
\]

### Step 2: Find the integrating factor
The integrating factor (IF) for a first-order linear differential equation of the form \(\frac{dy}{dt} + P(t) y = Q(t)\) is given by:

\[
\mu(t) = e^{\int P(t) dt}.
\]

Here, \(P(t) = \frac{6}{t}\), so we calculate the integrating factor:

\[
\mu(t) = e^{\int \frac{6}{t} dt} = e^{6 \ln t} = t^6.
\]

### Step 3: Multiply the differential equation by the integrating factor
Now multiply the entire equation by \(t^6\):

\[
t^6 \frac{dy}{dt} + t^6 \cdot \frac{6}{t} y = t^6 \left( 3 - \frac{2}{t} \right).
\]

Simplifying:

\[
t^6 \frac{dy}{dt} + 6t^5 y = 3t^6 - 2t^5.
\]

### Step 4: Solve the equation
Notice that the left-hand side is the derivative of \(t^6 y\):

\[
\frac{d}{dt} \left( t^6 y \right) = 3t^6 - 2t^5.
\]

Now integrate both sides with respect to \(t\):

\[
t^6 y = \int \left( 3t^6 - 2t^5 \right) dt.
\]

Integrating the right-hand side:

\[
t^6 y = \frac{3}{7} t^7 - \frac{2}{6} t^6 + C = \frac{3}{7} t^7 - \frac{1}{3} t^6 + C.
\]

### Step 5: Solve for \(y(t)\)
Now, solve for \(y(t)\):

\[
y(t) = \frac{1}{t^6} \left( \frac{3}{7} t^7 - \frac{1}{3} t^6 + C \right).
\]

Simplify:

\[
y(t) = \frac{3}{7} t - \frac{1}{3} + \frac{C}{t^6}.
\]

### Step 6: Apply the initial condition
We are given that \(y(1) = 0\). Substitute \(t = 1\) and \(y(1) = 0\) into the equation:

\[
0 = \frac{3}{7} \cdot 1 - \frac{1}{3} + \frac{C}{1^6}.
\]

Simplifying:

\[
0 = \frac{3}{7} - \frac{1}{3} + C.
\]

To solve for \(C\), find a common denominator between \(\frac{3}{7}\) and \(\frac{1}{3}\):

\[
0 = \frac{9}{21} - \frac{7}{21} + C = \frac{2}{21} + C.
\]

Thus:

\[
C = -\frac{2}{21}.
\]

### Step 7: Write the final solution
Substitute \(C = -\frac{2}{21}\) back into the solution for \(y(t)\):

\[
y(t) = \frac{3}{7} t - \frac{1}{3} - \frac{2}{21 t^6}.
\]

Thus, the solution to the initial-value problem is:

\[
y(t) = \frac{3}{7} t - \frac{1}{3} - \frac{2}{21 t^6}.
\]

Would you like further details or have any questions?

Here are five related questions to consider:
1. How do you find the integrating factor for a linear differential equation?
2. Why is it important to rewrite the differential equation in standard form before solving?
3. How does the initial condition influence the final solution of a differential equation?
4. What role does the integrating factor play in solving linear differential equations?
5. How would the solution change if the initial condition were different?

**Tip:** Always check that the initial condition satisfies your final solution to ensure correctness.

Give the solution y(t) of the initial-value problem t dy/dt + 6y = 3t - 2 with t > 0 and y(1) = 0.

Learn how to solve the first-order differential equation t dy/dt + 6y = 3t - 2 with the initial condition y(1) = 0. This solution covers the use of the integrating factor method and detailed steps to reach the final answer. Suitable for college-level calculus students.

Math Problem Statement

Give the solution y ( t ) y(t) of the initial-value problem t d y d t + 6 y

0 t>0 and y ( 1 )

y ( t )

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