Question

If $y=\frac{ax+b}{cx+d}$, then $2y_1{y}_3=$

• A. $3y_2^2$
• B. $y_2^2$
• C. $3y_1^2$
• D. $3y^2$

Answer 1

The correct option is A $3y_2^2$

$(cx+d)y=ax+b$
$cy+(cx+d)y^{{}^{\prime }}=a$
$c{y}^{{}^{\prime }}+c{y}^{{}^{\prime }}+(cx+d)y^{{}^{\prime \prime }}=0$
$2c{y}^{{}^{\prime \prime }1}+\left(c{y}^{{}^{\prime \prime }}\right)+(cx+d)y^{{}^{\prime \prime \prime }}=0$
$(cx+d)y^{{}^{\prime \prime \prime }}=-3c{y}^{{}^{\prime \prime }}$
$2\left(y^{{}^{\prime \prime \prime }}\right)\left(y^1\right)=-2\left(\frac{-3c{y}^{{}^{\prime \prime }}}{cx+d}\right)\frac{(cx+d)y^{{}^{\prime \prime }}}{2c}$
$=3(y^{{}^{\prime \prime }}{)}^2$