Solve ( D 2 − 3 D + 2 ) Y = 2 E X Sin ( X 2 ) - Applied Mathematics 2

Question

Solve $(D^{2} - 3 D + 2) y = 2 e^{x} \sin (\frac{x}{2})$

Solution

$(D^{2} - 3 D + 2) y = 2 e^{x} \sin (\frac{x}{2})$

For complementary function ,

$f (D) = 0$

∴ $(D^{2} - 3 D + 2) = 0$

Roots are : D=2,1 Real roots .

$y_{c} = c_{1} e^{x} + c_{2} e^{2} x$

For particular integral ,

$y_{p} = \frac{1}{f (D)} x$

= $\frac{1}{D^{2} - 3 D + 2} 2 e^{x} \sin (\frac{x}{2})$

= $2 e^{x} \frac{1}{{(D + 1)}^{2} - 3 (D + 1) + 2} \sin (\frac{x}{2})$

= $2 e^{x} \frac{1}{(D^{2} - D)} \sin (\frac{x}{2})$

= $2 e^{x} \frac{1}{- (\frac{1}{4}) - D} \sin (\frac{x}{2})$

= $- 8 e^{x} \frac{1}{4 D + 1} \sin (\frac{x}{2})$

= $- 8 e^{x} \frac{4 D - 1}{16 D^{2} - 1} \sin (\frac{x}{2})$

$y p = \frac{8}{5} e^{x} (- \sin (\frac{x}{2}) - 2 \cos (\frac{x}{2}))$ `

$y_{c} = y_{c} + y_{p} = c_{1} e^{x} + c_{2} e^{2 x} + \frac{8}{5} e^{x} (- \sin (\frac{x}{2}) - 2 \cos (\frac{x}{2}))$

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Linear Differential Equation with Constant Coefficient‐ Complementary Function

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Q 2.a Q 1.f Q 2.b

2016-2017 (June) CBCGS

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2016-2017 (June) CBCGS (with solutions)

Q 2.a | 6 marks

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Solve ( D 2 − 3 D + 2 ) Y = 2 E X Sin ( X 2 ) - Applied Mathematics 2

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