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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
Prove: 1/cos2A+sin2A/cos2A=sinA+cosA/cosA-sinA
A round balloon of radius 'a' subtends an angle θ at the eye of the observer while the angle of elevation of its centre is Φ.Prove that the height of the center of the balloon is a
44 breaths in 2 hours
Lindy has 48 chocolate chip cookies and 64 vanilla wafer cookies. How many bags can Lindy fill if she puts the chocolate chip cookies and the vanilla wafers in the same bag? She pl
Example determines the first four derivatives for following. y = cos x Solution: Again, let's just do so
conclusion on share and dividend project
Formulas
3x+3/x2 -6x+5
find the area of this figure in square millimeter measure each segment to the nearest millmeter
What is a way to solve indices
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