The second differential of the function is 3/7
Given the differential function [tex](x +2y)=\frac{dy}{dx} (2x-y)[/tex]
Differentiate both sides with respect to x implicitly
[tex](x +2y)=\frac{dy}{dx} (2x-y)[/tex]
[tex]1+2y'=y"(2x-y)+y'(2-y') \\y''=\frac{1+2y'}{(2x-y)+y'(2-y') }[/tex]
Get the first derivative at the point (3, 0)
[tex]y'=\frac{2x-y}{x+2y} \\y'=\frac{2(3)-0}{0+2(3)}\\y'=\frac{6}{6} \\y'=1[/tex]
Substitute into the second derivative to have:
[tex]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}[/tex]
Hence the second differential of the function is 3/7
learn more on differential equation here; https://brainly.com/question/18760518
Complete question:
If (x 2y)dy/dx=2x-y what is the value of [tex]\frac{d^2y}{dx^2}[/tex] the point (3,0)
