Question

If (x 2y)dy/dx=2x-y what is the value of.

Answer 1

The second differential of the function is 3/7

Given the differential function $(x +2y)=\frac{dy}{dx} (2x-y)$

Differentiate both sides with respect to x implicitly

$(x +2y)=\frac{dy}{dx} (2x-y)$

$1+2y'=y"(2x-y)+y'(2-y') \\y''=\frac{1+2y'}{(2x-y)+y'(2-y') }$

Get the first derivative at the point (3, 0)

$y'=\frac{2x-y}{x+2y} \\y'=\frac{2(3)-0}{0+2(3)}\\y'=\frac{6}{6} \\y'=1$

Substitute into the second derivative to have:

$y''=\frac{1+2y'}{(2x-y)+y'(2-y') }\\y''=\frac{1+2(1)}{(2(3)-0)+(1)(2-(1)) }\\y''=\frac{3}{6+1} \\y''=\frac{3}{7}$

Hence the second differential of the function is 3/7

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Complete question:

If (x 2y)dy/dx=2x-y what is the value of $\frac{d^2y}{dx^2}$ the point (3,0)