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CHHAYA PUBLICATION-LINEAR DIFFERENTIAL EQUATION -Short Answer Type Questions
- 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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- cos t dx/dt + sin t = 1
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- dy/dx + y sec x = tan x (0le x le pi/2)
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- (1 + x ) dy/dx - xy = 1 - x
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- (1-x^(2)) dy/dx - xy = x^(2) , given y = 2 when x = 0
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- dx/dt + x cos t = 1/2 sin 2 t
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- dy/dx + (tan x + 1/x ) y = (sec x )/x
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- dy/dx + 2y tan x = sinx, give y = 0 when x= pi/3
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- dy/dx= (xsqrt(x^(2) - 1) -y )/sqrt(x^(2)-1), given y = 1 when x = 1
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- x(1-x^(2)) dy + (2x^(2)y-y-5x^(3)) dx = 0
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- x(dy/dx + y ) = 1 -y
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- (x + 2y^(3)) dy/dx = y
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- ydx - (x + 2y ^(2) ) dy = 0
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- 1 + y^(2) + (x - e^(tan^(-1)y)) dy/dx = 0
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- 1 + y^(2) +(x-e^(-tan ^(-1)y)) dy/dx=0
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- y^(2) + (x - 1/y ) dy/ dx = 0
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- dx+ x dy = e^(-y) sec ^(2) y dy
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- (x^(2) y^(3) + 2xy) dy = dx
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