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There actually isn't a whole lot to do throughout this case. We'll find two solutions which will form a basic set of solutions and therefore our general solution will be as,
Example: Solve the following initial value problem
2 x2 y′′ + 3xy′ -15 y = 0,
y (1) = 0
y′ (1) = 1
Solution
We first require finding the roots to (3).
2r ( r -1) + 3r -15 = 0
2r2 + r -15 = (2r - 5)(r + 3) = 0
⇒ r1 = 5/2 and r2 = -3
Then the general solution is,
y(x) = c1x5/2 + c2 x-3
To get the constants we differentiate and plug into the initial conditions where we did back into the second order differential equations section.
y'(x) = (5/2) c1x3/2 - 3c2 x-4
0 = y(1) = c1 + c2
1 = y'(1) = (5/2)c1 + (-3) c2
By solving these equations we get:
c1 = 2/11,
c2 = -(2/11)
the actual solution is,
y(x) = (2/11) x5/2 -(2/11) c2 x-3
There's a nice way to show why the expresion for the area of a circle of radius R is: Pi * R 2 . It has an comman relationship with the experation for the circumference of a
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We contain a piece of cardboard i.e. 14 inches by 10 inches & we're going to cut out the corners as illustrates below and fold up the sides to form a box, also illustrated below. F
x^2 + 10x +2 4
The base of a right cylinder is the circle in the xy -plane with centre O and radius 3 units. A wedge is obtained by cutting this cylinder with the plane through the y -axis in
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