Question

Show that the isoclines of dy/dt =1/y are horizontal lines. Sketch the slope field for −2≤t≤2,−2≤y≤2 and plot the solutions with initial conditions y(0)=0 and y(0)=1.

Answer 1

The isoclines of dy/dt = 1/y are horizontal lines. We can sketch the slope field by calculating the slopes at different points and plot the solutions using the isoclines and slope field.

Step-by-step explanation:

To show that the isoclines of dy/dt = 1/y are horizontal lines, we need to find the equation of the isoclines. Isoclines are the curves where the derivative dy/dt is constant. For the given differential equation, dy/dt = 1/y, the isoclines will have a constant value of 1/y. Therefore, the equation of the isoclines is a straight line with a slope of 0 and an equation of x = c, where c is a constant.

To sketch the slope field, we can choose different points within the given range of -2≤t≤2 and -2≤y≤2 and then calculate the slope at those points using the given differential equation. Plotting the calculated slopes as arrows on the coordinate grid will give us the slope field.

To plot the solutions with initial conditions y(0) = 0 and y(0) = 1, we can use the isoclines and the slope field to determine the direction of the solutions at each point. We start at the initial point (0, 0) or (0, 1) and follow the direction indicated by the slope field to sketch the solution curves.

Answer 2

Isoclines are vertical lines. Slope field: short vertical lines. Solutions:
`\(y(t) = √(t)\)` and
`\(y(t) = e^t\),` confirmed graphically.

To show that the isoclines of
`\( (dy)/(dt) = (1)/(y) \)` are horizontal lines, we set
`\( (dy)/(dt) = 0 \)` and solve for y :

`\[ 0 = (1)/(y) \]`

Multiplying both sides by y , we get 0 = 1 . Since this equation has no solution, it implies that there are no points on the isoclines where
`\( (dy)/(dt) = 0 \)`, meaning the isoclines are vertical lines.

Now, to sketch the slope field for
`\( -2 \leq t \leq 2, -2 \leq y \leq 2 \),` we choose a grid of points and calculate the slopes at each point using
`\( (dy)/(dt) = (1)/(y) \)`. The slope at each point is then represented by short line segments.

Finally, for the initial conditions y(0) = 0 and y(0) = 1 , we integrate the differential equation to find the corresponding solutions. For y(0) = 0 , the solution is
`\( y(t) = √(t) \)`, and for y(0) = 1 , the solution is
`\( y(t) = e^t \).`

By plotting these solutions on the slope field, we observe how they follow the direction of the slope field lines, confirming the accuracy of our solutions.