Question

In parts (a) and (b) sketch isoclines(see the Remarks on page 39) for the given differential equation using the indicated values of. Construct a direction field over a grid by carefully drawing lineal elements with the appropriate slope at chosen points on each isocline. In each case, use this rough direction field to sketch an approximate solution curve for the IVP consisting of the DE and the initial condition.

(a); an integer satisfying.

(b);,,.

Short answer

a).Isoclines for the given equation are lines of the form\(x + y = c\).

b). Isoclines for the given equation are lines of the form\({x^2} + {y^2} = c\).

Step-by-step solution

Direction field.

If we systematically evaluate \(f\) over a rectangular grid of points in the \(xy\)-plane and draw a line element at each point \((x,y)\)of the grid with slope \(f(x,y),\)then the collection of all these line elements is called a direction field or a slope field of the differential equation \(\frac{{dy}}{{dx}} = f(x,y)\).

Solution for part a).

The given differential equation, can be written as,

\(\frac{{dy}}{{dx}} = - \frac{x}{y}\)

(a).

Isoclines for the given equation are lines of the form \((x + y = c)\) where \(c\) is substituted for some integers between \( - 5\) and\(5\).

For \(c = - 5, - 4, - 3, - 2, - 1,0,1,2,3,4,5\), the isoclines should look like this, shown below:

Now we sketch the direction field. Elements that lie on the same line have the same slope. Their slopes will be given by the value of \(c\)

The sketch of the directional field is shown below,

Finally, we sketch the solution passing through \((0,1)\)by connecting the elements in the direction field.

Solution for part b).

Isoclines for the given equation are lines of the form \(({x^2} + {y^2} = c)\) where \(c\) is substituted for integers \(\frac{1}{4},1,\frac{9}{4}\) and\(4\).

For\(c = \frac{1}{4},1,\frac{9}{4},4\), the isoclines should look like this, shown below:

Now we sketch the direction field. Elements that lie on the same line have the same slope. Their slopes will be given by the value of \(c\).

Finally, we sketch the solution passing through\((0,1)\)by connecting the elements in the direction field.

Therefore, Isoclines for the given equation are lines of the form \(x + y = c\) or\({x^2} + {y^2} = c\).