If x (d2 y/d x2)+(d y/d x)= ln x, y(1)=1 and y prime(1)=-1 then y prime prime((1/3)) equals

Question

If x (d2 y/d x2)+(d y/d x)= ln x, y(1)=1 and y prime(1)=-1 then y prime prime((1/3)) equals (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Given ( d / dx )( x ( dy / dx ))= ln x ⇒ x ( dy / dx )= x ln x - x + C Now y prime(1)=-1, so -1=0-1+ C ⇒ C =0 ∴ ( dy / dx )= ln x -1 Hence .( d 2 y / dx 2)] x =(1/3)=(1/ x )=3

Q. If $x \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = ln x, y (1) = 1$ and $y^{'} (1) = - 1$ then $y^{''} (\frac{1}{3})$ equals

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Given $\frac{d}{d x} (x \frac{d y}{d x}) = ln x$
$\Rightarrow x \frac{d y}{d x} = x ln x - x + C$
Now $y^{'} (1) = - 1$ , so
$- 1 = 0 - 1 + C \Rightarrow C = 0$
$∴ \frac{d y}{d x} = ln x - 1$
Hence $\frac{d ^{2} y}{d x ^{2}}]_{x = \frac{1}{3}} = \frac{1}{x} = 3$

Q. If xdx2d2y​+dxdy​=lnx,y(1)=1 and y′(1)=−1 then y′′(31​) equals

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Given dxd​(xdxdy​)=lnx ⇒xdxdy​=xlnx−x+C Now y′(1)=−1, so −1=0−1+C⇒C=0 ∴dxdy​=lnx−1 Hence dx2d2y​]x=31​​=x1​=3

Q. If $x \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = ln x, y (1) = 1$ and $y^{'} (1) = - 1$ then $y^{''} (\frac{1}{3})$ equals

Given $\frac{d}{d x} (x \frac{d y}{d x}) = ln x$
$\Rightarrow x \frac{d y}{d x} = x ln x - x + C$
Now $y^{'} (1) = - 1$ , so
$- 1 = 0 - 1 + C \Rightarrow C = 0$
$∴ \frac{d y}{d x} = ln x - 1$
Hence $\frac{d ^{2} y}{d x ^{2}}]_{x = \frac{1}{3}} = \frac{1}{x} = 3$