Solve the following initial-value problems with the initial condition \( y_{0}=0 \) and graph the solution. 119. \( \frac{d y}{d t}=y+1 \) 120. \( \frac{d y}{d t}=y-1 \) 121. \( \frac{d y}{d t}=y+1 \)

To solve the given initial-value problems, we can start by separating variables, integrating, and then applying the initial condition \( y(0) = 0 \). 1. For the equation \( \frac{d y}{d t}=y+1 \): \[ \frac{dy}{y+1} = dt \] Integrating both sides: \[ \ln|y+1| = t + C \] Exponentiating gives us: \[ y+1 = e^{t+C} = e^t \cdot e^C \] Letting \( K = e^C \), we find: \[ y = Ke^t - 1 \] Now apply the initial condition \( y(0) = 0 \): \[ 0 = Ke^0 - 1 \implies K = 1 \] So, the solution is: \[ y(t) = e^t - 1 \] 2. For the equation \( \frac{d y}{d t}=y-1 \): \[ \frac{dy}{y-1} = dt \] Integrating both sides: \[ \ln|y-1| = t + C \] Exponentiating gives us: \[ y-1 = e^{t+C} = e^t \cdot e^C \] Letting \( K = e^C \): \[ y = Ke^t + 1 \] Applying the initial condition \( y(0) = 0 \): \[ 0 = Ke^0 + 1 \implies K = -1 \] Thus, the solution is: \[ y(t) = -e^t + 1 \] 3. The equation \( \frac{d y}{d t}=y+1 \) is the same as the first problem, so we can reuse the solution: \[ y(t) = e^t - 1 \] Now, let's summarize the solutions: - For \( \frac{d y}{d t}=y+1 \), the solution is \( y(t) = e^t - 1 \). - For \( \frac{d y}{d t}=y-1 \), the solution is \( y(t) = -e^t + 1 \). - The solution is the same for the last equation, \( y(t) = e^t - 1 \). To graph these solutions, you would have one increasing curve starting from \( (0, -1) \) for the first and third problems, and one decreasing curve starting from \( (0, 0) \) for the second. Enjoy plotting!