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We're here going to take a brief detour and notice solutions to non-constant coefficient, second order differential equations of the form.
p (t) y′′ + q (t ) y′ + r (t ) y = 0
Generally, finding solutions to these types of differential equations can be much tougher than determining solutions to constant coefficient differential equations. Though, if we previously know one solution to the differential equation we can utilize the method that we used in the previous section to find a second solution. This method is termed as reduction of order.
The vertices of a ? ABC are A(4, 6), B(1. 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4 .Calculate the ar
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2(sin 6 ?+cos 6 ?) - 3(sin 4 ?+cos 4 ?)+1 = 0 Ans: (Sin 2 ?)3 + (Cos 2 ?)3-3 (Sin 4 ?+(Cos 4 ?)+1=0 Consider (Sin 2 ?)3 +(Cos 2 ?)3 ⇒(Sin 2 ?+Cos 2 ?)3-3 Sin 2 ?Co
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