The D.E whose solution is $y = A cos 2 x + B sin 2 x + C$ given by:

y_{3} - 4 y_{1} + 2 y = 0

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y_{3} + 4 y_{1} = 0

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y_{3} + 4 y_{1} + 2 = 0

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y_{3} + y_{2} + 2 y = 0

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Solution

The correct option is D $y_{3} + 4 y_{1} = 0$
$y = A c o s 2 x + B s i n 2 x + C$
$y_{1} = - 2 A s i n 2 x + 2 B c o s 2 x + 0$
$y_{2} - 4 A c o s 2 x - 4 B s i n 2 x$
$\Rightarrow y_{2} = - 4 (y - c)$
$\Rightarrow y_{2} + 4 y = 4 c$
$\Rightarrow y_{3} + 4 y_{1} = 0$

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Similar questions

Q. If

y = e^{2 x} (a x + b)

, show that

y_{2} - 4 y_{1} + 4 y = 0

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(x_{1}, y_{1}), (x_{2}, y_{2})

and

(x_{3}, y_{3})

are collinear, show that

\sum (\frac{y_{1} - y_{2}}{x_{1} x_{2}}) = 0

, i.e

\frac{y_{1} - y_{2}}{x_{1} x_{2}} + \frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} = 0

Q. The solution of

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If three points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ lie on the same line, prove that $\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0.$

Find the sum of the following polynomials

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The D.E whose solution is y=Acos2x+Bsin2x+C given by:

The correct option is D y3+4y1=0y=Acos2x+Bsin2x+Cy1=−2Asin2x+2Bcos2x+0y2−4Acos2x−4Bsin2x⇒y2=−4(y−c)⇒y2+4y=4c⇒y3+4y1=0

The D.E whose solution is $y = A cos 2 x + B sin 2 x + C$ given by:

The correct option is D $y_{3} + 4 y_{1} = 0$
$y = A c o s 2 x + B s i n 2 x + C$
$y_{1} = - 2 A s i n 2 x + 2 B c o s 2 x + 0$
$y_{2} - 4 A c o s 2 x - 4 B s i n 2 x$
$\Rightarrow y_{2} = - 4 (y - c)$
$\Rightarrow y_{2} + 4 y = 4 c$
$\Rightarrow y_{3} + 4 y_{1} = 0$