Math, asked by krishnakumarid1969, 6 months ago

solve d³y/dx³-3dy/dx+2y=0​

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Answered by aayushp2016
0

Answer:

Solution (i) Let take f(x) = x3 - 2x2 - x + 2 The constant term in f(x) is are ±1 and ±2 Putting x = 1 in f(x), we have f(1) = (1)3 - 2(1)2 -1 + 2 = 1 - 2 - 1 + 2 = 0 According to remainder theorem f(1) = 0 so that (x - 1) is a factor of x3 - 2x2 - x + 2 Putting x = - 1 in f(x), we have f(-1) = (-1)3 - 2(-1)2 –(-1) + 2 = -1 - 2 + 1 + 2 = 0 According to remainder theorem f(-1) = 0 so that (x + 1) is a factor of x3 - 2x2 - x + 2 Putting x = 2 in f(x), we have f(2) = (2)3 - 2(2)2 –(2) + 2 = 8 -82 - 2 + 2 = 0 According to remainder theorem f(2) = 0 so that (x – 2 ) is a factor of x3 - 2x2 - x + 2 Here maximum power of x is 3 so that its can have maximum 3 factors

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Answered by Chidagni
0

Answer:

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