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CHHAYA PUBLICATION-LINEAR DIFFERENTIAL EQUATION -Short Answer Type Questions
- dy/dx - 2/(x+1)y = (x+1)^(3)
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- dy/dx - ytan x = - 2 sin x
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- dy/dx + y cot x = 2 cos x
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- (x-x^(3)) dy/dx + (2x^(2)-1) y = a x^(3)
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- dy/dx- y tan x = e^(2)
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- (x^(2)+ 1) dy/dx+ 2xy = 4x^(2)
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- xdy/dx + 2y = log x
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- Solve (dy)/(dx)+ 3y = e^(-2x)
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- cos^(2) x dy/dx + y = tanx [0le x le pi/2]
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- dy/dx - n/xy = e^(x)x^(n)
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- dy/dx - y tan x = e^(x) sec x
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- dy/dx + n/x y = a/x^(n)
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- dy/dx - 3y cot x = sin 2x, given y= 2 when x = pi/ 2
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- dy/dx +2y = 6e^(2)
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- dy/dx + (2x)/(1+ x^(2)) y = 1/((1+x^(2))^(2))
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- 2 cosx dy/dx + 2 ysin x = sin 2 x
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- dy/dx + (4x)/(x^(2) + 1) y = 1/((x^(2) + 1)^(3))
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- dy/dx + 1/(xlog x ) y = 2/x
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- (x^(2) -1) dy/dx + 2xy = 2/(x^(2)-1)
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- (1+x^(2)) dy/dx + 2xy = 4x^(2), given y = 0 , when x = 0
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