Question

slove (x2d2-xd-3)y=x2 log x

Answer 1

Answer:

$solve\:for\:d,\:\left(x^2d^2-xd-3\right)y=x^2\log \left(x\right)\quad :\quad d=\frac{y+\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy},\:d=\frac{y-\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy} ; x^2y\neq 0$

Step-by-step explanation:

$\left(x^2d^2-xd-3\right)y=x^2\log \left(x\right)$

$d^2x^2y-dxy-3y=x^2\log \left(x\right)$

$\mathrm{Subtract\:}x^2\log \left(x\right)\mathrm{\:from\:both\:sides}$

$d^2x^2y-dxy-3y-x^2\log \left(x\right)=x^2\log \left(x\right)-x^2\log \left(x\right)$

$x^2yd^2-xyd-3y-x^2\log \left(x\right)=0$

$d_{1,\:2}=\frac{-\left(-xy\right)\pm \sqrt{\left(-xy\right)^2-4x^2y\left(-3y-x^2\log \left(x\right)\right)}}{2x^2y}$

$d_{1,\:2}=\frac{-\left(-xy\right)\pm \:x\sqrt{y\left(13y+4x^2\log \left(x\right)\right)}}{2x^2y};\quad \:x^2y\ne \:0$

$\frac{-\left(-xy\right)+x\sqrt{y\left(13y+4x^2\log \left(x\right)\right)}}{2x^2y}$

$=\frac{xy+x\sqrt{y\left(13y+4x^2\log \left(x\right)\right)}}{2x^2y}$

$=\frac{x\left(y+\sqrt{\left(13y+x^2\cdot \:4\log \left(x\right)\right)y}\right)}{2x^2y}$

$=\frac{y+\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy}$

$\frac{-\left(-xy\right)-x\sqrt{y\left(13y+4x^2\log \left(x\right)\right)}}{2x^2y}$

$=\frac{xy-x\sqrt{y\left(13y+4x^2\log \left(x\right)\right)}}{2x^2y}$

$=\frac{x\left(y-\sqrt{\left(13y+x^2\cdot \:4\log \left(x\right)\right)y}\right)}{2x^2y}$

$=\frac{y-\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy}$

$d=\frac{y+\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy},\:d=\frac{y-\sqrt{y\left(4x^2\log \left(x\right)+13y\right)}}{2xy};\quad \:x^2y\ne \:0$