# 演習問題

$\frac{dy}{dx}=2$

$\frac{dy}{dx}=x$

$\frac{dy}{dx}=3y$

$\frac{dy}{dx}=-2y$

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

$\frac{dy}{dx}=\frac{1+y}{1+x}$

$\frac{dy}{dx}=2y+3$

$\frac{dy}{dx}=3{x}^{2}y$

$\frac{dy}{dx}=4{x}^{2}$

$\frac{dy}{dx}=3x+1$

$\frac{dy}{dx}=5{x}^{3}-7{x}^{2}-x+5$

$\frac{dy}{dx}=xy$

$\frac{dy}{dx}=\frac{y}{x}$

$\frac{dy}{dx}=\mathrm{sin}x$

$\frac{dy}{dx}=\mathrm{cos}x$

$\frac{dy}{dx}=4\mathrm{sin}5x$

$\frac{dy}{dx}=6\mathrm{cos}2x$

$\frac{dy}{dx}=2{e}^{3x}$

$\frac{dy}{dx}=-\frac{\mathrm{cos}x}{\mathrm{sin}y}$

$\frac{dy}{dx}=\frac{1+{e}^{x}}{{e}^{x-y}}$

$\left({\mathrm{tan}}^{-1}y\right){y}^{\prime }=\frac{\sqrt{2-{x}^{2}}}{{x}^{2}-2}$

${y}^{2}dx-{x}^{3}dy=0$

$\sqrt{x-1}{y}^{\prime }=\frac{1}{\sqrt{y}}$

${y}^{\prime }=\frac{x{y}^{3}+x{y}^{2}}{{x}^{3}y-xy}$

${y}^{\prime }={sin}^{2}x\mathrm{sin}y\mathrm{cos}x$

$\frac{dy}{dx}=\frac{{e}^{y}}{{e}^{x}}$

$\left({y}^{2}+\mathrm{sin}y\right){y}^{\prime }+\mathrm{cos}x+{x}^{3}=0$

$\frac{dy}{dx}=\frac{1}{2xy}$

($x=1$,$y=1$ )

$\frac{dy}{dx}={e}^{2x}{e}^{3y}$      （$x=0$,$y=0$

$\frac{dy}{dx}=\mathrm{sin}x{cos}^{2}y$      ($x=\frac{\pi }{4}$,$y=0$)

$\frac{dy}{dx}=\frac{1+{y}^{2}}{\sqrt{1-{x}^{2}}}$      ($y=\sqrt{3}$,$x=\frac{1}{2}$)