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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. $y=a\: e^{3x}+b\: e^{-2x}$

$\begin{array}{1 1} (A)\;y'' - y' - 6y = 0. \\(B)\;y'' - y' - y = 0. \\ (C)\;y'' - y - 6y' = 0. \\ (D)\;y''' - y' - 6y = 0. \end{array} $

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1 Answer

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Toolbox:
  • After differentiating, to eliminate the arbitrary constants we can convert the equations, into a determinant form and expand along their rows or columns to eliminate the arbitrary constants.
Step 1:
Given : $y = ae^3x + be^{-2x}$ -----------(1)
Differentiate on both sides
$y' 3ae^{3x} - 2be^{-2x}$-------------(2)
Again differentate equ (2) on both sides
$y'' = 9ae^{3x} + 4be^{-2x}$-----------(3)
Step 2:
As the hint in the tool box says, we can write the three equations in the form a determinant to eliminate the arbitrary constants
$\begin{vmatrix}y&y'&y''\\1&3&9\\1&-2&4\end{vmatrix}=0$
Step 3:
On expanding the determinant along the rows we get
$y(12 + 18) - y'(4 - 9) + y''(4-9) = 0$
$30y +5y' -5y'' = 0$
Dividing throughout by -5 we get,
$y'' - y' - 6y = 0.$
answered Aug 16, 2013 by sreemathi.v
 

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