Question

lf $\log y=ta{n}^{-1}x$ then $(1+x^2)y_2+(2x-1)y_1=$

• A. 0
• B. 1/2
• C. -1
• D. 1

Answer 1

The correct option is A 0

$\frac{1}{y}\left(y^{{}^{\prime }}\right)=\frac{1}{1+x^2}$
$y^{{}^{\prime }}=\frac{y}{1+x^2}$
$y^{{}^{\prime \prime }}=\frac{y^{\prime }}{1+x^2}-y\frac{2x}{(1+x^2{)}^2}$
$(1+x^2)y^{{}^{\prime \prime }}=y^{{}^{\prime }}-y\frac{2x}{(1+x^2)}$
$=y^{{}^{\prime }}-y^{{}^{\prime }}\left(2x\right)$
$(1+x^2)y^{{}^{\prime \prime }}=(1-2x)y^{{}^{\prime }}$
$(1+x^2)y^{{}^{\prime \prime }}+(2x-1)y^{{}^{\prime }}=0$