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Formation of a Differential Equation from a General Solution

Match the col...

Question

Match the column
(Note : $y_{n}$ represents the $n^{t h}$ derivative wrt. $x$ )

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Solution

A) $y = a e^{3 x} + b e^{5 x} \Rightarrow y_{1} = 3 a e^{3 x} + 5 b e^{5 x} \Rightarrow y_{1} = 3 y + 2 b e^{5 x} \Rightarrow y_{2} = 3 y_{1} + 5 (y_{1} - 3 y) \Rightarrow y_{2} - 8 y_{1} + 15 y = 0$
B) $x y = a e^{x} + b e^{- x} + x^{2} \Rightarrow x y_{1} + y = a e^{x} - b e^{- x} + 2 x \Rightarrow x y_{2} + 2 y_{1} = a e^{x} + b e^{- x} + 2 \Rightarrow x y_{2} + 2 y_{1} = x y - x^{2} + 2 \Rightarrow x y_{2} + 2 y_{1} - x y + x^{2} - 2 = 0$
C) $a x^{2} + b y^{2} = 1 \Rightarrow 2 a x + 2 b y y_{1} = 0 \Rightarrow a x + b y y_{1} = 0$ ...(1)
$\Rightarrow a + b {(y_{1})}_{1}^{2} + b y y_{2} = 0 \Rightarrow a + b [{(y_{1})}_{1}^{2} + y y_{2}] = 0$ ...(2)
Multiplying (2) by x and subtracting from (1), we get
$b x [y y_{2} + {y_{1}}^{2}] - b y y_{1} = 0 \Rightarrow x {y y_{2} + {(y_{1})}_{1}^{2}} = y y_{1}$
D) $y = a sin (4 x + b) \Rightarrow y_{1} = 4 a cos (4 x + b) \Rightarrow y_{2} = - 16 sin (4 x + b) \Rightarrow y_{2} = - 16 y$

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