If curve satisfying x(x+1)y1−y=x(x+1) passes through (1,0), then the value of 54y(4)−log4 is
A
3
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B
2
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C
1
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D
0
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Solution
The correct option is A 3 x(x+1)dydx−y=x(x+1)⇒dydx−yx2+x=1 ...(1) Let u=e∫−dxx2+x=1x+1 Multiplying both sides of (1) by u, we get (1x+1)dydx−(1x+1)yx2+x=(1x+1)⇒(1x+1)dydx+ddx(1x+1)y=(1x+1) Using gdfdx+fdgdx=d(fg)dx ddx((1x+1)y)=(1x+1) Integrating both sides ∫ddx((1x+1)y)dx=∫(1x+1)dx⇒(1x+1)y=x+logx+c As it passes through (1,0), we get c=−1 Now for x=4 (14+1)y(4)=4+log4−1⇒54y(4)−log4=3