Solve the following initial value problem: \[ t \frac{d y}{d t}+4 y=2 t \] with \( y(1)=3 \). Put the problem in standard form. Then find the integrating factor, \( \rho(t)= \) Next item and finally find \( y(t)= \) Submit answer

To solve the initial value problem \[ t \frac{d y}{d t} + 4 y = 2 t \] with the initial condition \( y(1) = 3 \), we will follow these steps: ### Step 1: Put the problem in standard form We start by rewriting the equation in standard linear form, which is \[ \frac{dy}{dt} + P(t) y = Q(t) \] To do this, we divide the entire equation by \( t \): \[ \frac{dy}{dt} + \frac{4}{t} y = 2 \] Here, we have: - \( P(t) = \frac{4}{t} \) - \( Q(t) = 2 \) ### Step 2: Find the integrating factor The integrating factor \( \rho(t) \) is given by \[ \rho(t) = e^{\int P(t) dt} = e^{\int \frac{4}{t} dt} \] Calculating the integral: \[ \int \frac{4}{t} dt = 4 \ln |t| = \ln |t|^4 \] Thus, the integrating factor is: \[ \rho(t) = e^{\ln |t|^4} = |t|^4 \] Since \( t > 0 \) in our context, we can simplify this to: \[ \rho(t) = t^4 \] ### Step 3: Multiply the equation by the integrating factor Now we multiply the entire differential equation by \( t^4 \): \[ t^4 \frac{dy}{dt} + 4 t^3 y = 2 t^4 \] ### Step 4: Rewrite the left-hand side The left-hand side can be expressed as the derivative of a product: \[ \frac{d}{dt}(t^4 y) = 2 t^4 \] ### Step 5: Integrate both sides Now we integrate both sides with respect to \( t \): \[ \int \frac{d}{dt}(t^4 y) dt = \int 2 t^4 dt \] This gives us: \[ t^4 y = \frac{2}{5} t^5 + C \] where \( C \) is the constant of integration. ### Step 6: Solve for \( y \) Now, we solve for \( y \): \[ y = \frac{2}{5} t + \frac{C}{t^4} \] ### Step 7: Apply the initial condition We use the initial condition \( y(1) = 3 \): \[ 3 = \frac{2}{5}(1) + \frac{C}{1^4} \] This simplifies to: \[ 3 = \frac{2}{5} + C \] Solving for \( C \): \[ C = 3 - \frac{2}{5} = \frac{15}{5} - \frac{2}{5} = \frac{13}{5} \] ### Step 8: Write the final solution Substituting \( C \) back into the equation for \( y \): \[ y = \frac{2}{5} t + \frac{13/5}{t^4} \] Thus, the final solution is: \[ y(t) = \frac{2}{5} t + \frac{13}{5 t^4} \] ### Summary of Results - The integrating factor is \( \rho(t) = t^4 \). - The solution to the initial value problem is \[ y(t) = \frac{2}{5} t + \frac{13}{5 t^4} \]