If x d 2 y d x 2 - d y d x - - n x - y 1 - 1 and y - 1 - - 1 then y - 1 3 equals

The correct option is A 1 xd^2ydx^2+dydx=ln x , #160; #160; y( 1)=1 #160; #160; #160; #160; #160; #160; #160; #160 ...

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Combination

If x d 2 y...

Question

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

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Similar questions

\frac{d^{2} y}{d x^{2}} = 2 x y

x \frac{d^{2} y}{d x^{2}} = y_{1}

x \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} + y = 0

x \frac{d^{2} y}{d x^{2}} = 2 x y

\frac{d y}{d x} = y \tan x, y (0) = 1

2 x \frac{d y}{d x} = 5 y, y (1) = 1

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

\frac{d y}{d x} = 2 x y, y (0) = 1

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

2 (y + 3) - x y \frac{d y}{d x} = 0

\frac{d y}{d x} = y \tan x, y (0) = 1

2 x \frac{d y}{d x} = 5 y, y (1) = 1

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

\frac{d y}{d x} = 2 x y, y (0) = 1

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}} = 1

\frac{d y}{d x} = - \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

Multiply x^{2} + 2 y by x^{3} - 2 x y + y^{3} and find the value of the product for x = 1 and y = - 1.

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