Proceed as in example 6 in section 2.3 to solve the given initial-value problem. dy dx + 2xy = f(x), y(0) = 6, where f(x) = x, 0 ≤ x < 1 0, x ≥ 1

The given initial value problem is

It is given that

y(0) = 6, where f(x) = x, 0 ≤ x < 1 0, x ≥ 1

Now the equation reduces to

 -------------(1)

Integrating factor=

Multiplying equation (1), both sides by integrating factor i.e

The equation reduces to

Integrating both sides, the equation reduces to

Now integrating the function which is on right hand side, we get

put and then differentiating,we get

2 x dx = dt

⇒x dx = dt/2

                                =

Replacing t by x², we get

   ---------------(2)

It is given that when x=0,y=6.

Putting these value in equation (2),we get

6=1/2 + C

⇒C= 6-1/2

⇒ C =11/2

The solution of the initial value problem is

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.