Let $y(t)$ represent your bank account balance, in dollars, after $t$ years. Suppose you start with $\$ 50000$ in the account. Each year the account earns $8 \%$ interest, and you deposit $\$ 4000$ into the account. This can be modeled with the differential equation: \[ \begin{array}{l} \frac{d y}{d t}=0.08 y+4000 \\ y(0)=50000 \end{array} \] Solve this differential equation for $y(t)$

Step 1 ：Rewrite the differential equation in the form \(\frac{dy}{dt} = f(t)g(y)\):

Step 2 ：\(\frac{dy}{dt} = 0.08y + 4000\)

Step 3 ：Separate the variables:

Step 4 ：\(\frac{1}{0.08y + 4000}dy = dt\)

Step 5 ：Integrate both sides:

Step 6 ：\(\int \frac{1}{0.08y + 4000}dy = \int dt\)

Step 7 ：Evaluate the integrals:

Step 8 ：\(\ln|0.08y + 4000| = t + C_1\)

Step 9 ：Solve for \(y\):

Step 10 ：\(0.08y + 4000 = C_2e^{t}\)

Step 11 ：\(y = \frac{C_2e^{t} - 4000}{0.08}\)

Step 12 ：Apply the initial condition \(y(0) = 50000\):

Step 13 ：\(50000 = \frac{C_2e^{0} - 4000}{0.08}\)

Step 14 ：\(C_2 = 8000\)

Step 15 ：Substitute \(C_2\) back into the equation for \(y(t)\):

Step 16 ：\(y = \frac{8000e^{t} - 4000}{0.08}\)

Step 17 ：The solution to the given differential equation is \(\boxed{y = \frac{8000e^{t} - 4000}{0.08}}\)