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. \( \frac{d y}{d t}=y+1 \)

**Problem 119** We are given \[ \frac{dy}{dt} = y+1,\quad y(0) = 0. \] Separate variables: \[ \frac{dy}{y+1} = dt. \] Integrate both sides: \[ \int \frac{dy}{y+1} = \int dt \quad\Longrightarrow\quad \ln|y+1| = t + C. \] Exponentiate: \[ |y+1| = e^{t+C} = Ae^t,\quad \text{where } A=e^C>0. \] Since \( y(0)=0 \) we have \[ |0+1| = A e^0 \quad\Longrightarrow\quad 1 = A, \] and because \( y+1 \) is positive near \( t=0 \), we take \[ y+1=e^t. \] Thus, the solution is \[ y = e^t - 1. \] **Graph Description for Problem 119** The graph of \( y = e^t - 1 \) starts at \( (0,0) \) and increases exponentially as \( t \) increases. As \( t\to -\infty \), \( y\to -1 \) (a horizontal asymptote). --- **Problem 120** We are given \[ \frac{dy}{dt}= y-1,\quad y(0)=0. \] Separate variables: \[ \frac{dy}{y-1} = dt. \] Integrate: \[ \int \frac{dy}{y-1} = \int dt \quad\Longrightarrow\quad \ln|y-1| = t + C. \] Exponentiate: \[ |y-1| = Ae^t,\quad \text{with } A=e^C>0. \] Use the initial condition \( y(0)=0 \): \[ |0-1| = A e^0 \quad\Longrightarrow\quad 1 = A. \] Since \( y(0)-1=-1 \) is negative and we want continuity, we choose \[ y-1=-e^t. \] Thus, the solution is \[ y= 1-e^t. \] **Graph Description for Problem 120** The graph of \( y = 1-e^t \) starts at \( (0,0) \) and decreases exponentially as \( t \) increases. As \( t\to \infty \), \( y\to -\infty \) and as \( t\to -\infty \), \( y\to 1 \) (a horizontal asymptote). --- **Problem 121** We are given \[ \frac{dy}{dt}= y+1,\quad y(0)=0. \] This is identical to Problem 119. Following the same steps, the solution is: \[ y = e^t - 1. \] **Graph Description for Problem 121** The graph is the same as for Problem 119: it starts at \( (0,0) \), increases exponentially as \( t \) increases, and has a horizontal asymptote at \( y=-1 \) as \( t\to -\infty \).