Differential Equations Set-2

$y=\dfrac{x^3}{3}+Cx$

$y=\dfrac{x^3}{2}+Cx$

$y=\dfrac{x^3}{2}+\dfrac{C}{x}$

$y=\dfrac{x^3}{2}+Cx^2$

$y=\dfrac{1}{x^2(x-2)}$

$y=\dfrac{1}{2x^2}$

$y=\dfrac{1}{x(2-x)}$

$y=\dfrac{1}{x^2(2-x)}$

$y=Cx^{-3}$

$y=Cx^{3}$

$y=\dfrac{C}{x^{2}}$

$y=Cx^{2}$

(I) $y=Cx+C^3$; (II) $y^2-\dfrac{4}{27}x^3=0$

(I) $y=Cx-C^3$; (II) $y^2+\dfrac{4}{27}x^3=0$

(I) $y=Cx+C^3$; (II) $y^2+\dfrac{4}{27}x^3=0$

(I) $y=Cx+C^3$; (II) $x=3t^2,\;y=2t^3$

$y=C_1x^2+C_2x$

$y=C_1x+C_2x^2\ln x$

$y=C_1x^2+C_2x^{-2}$

$y=C_1x^2+C_2x^2\ln x$

$y=e^{x}$

$y=e^{-x}+e^{-2x}$

$y=e^{-x}$

$y=e^{-x}-e^{-2x}$

$4\arctan\!\left(\dfrac{y}{x}\right)-\ln\!\left(x^{2}+y^{2}\right)=\pi-\ln 2$

$4\arctan\!\left(\dfrac{y}{x}\right)-\ln\!\left(x^{2}+y^{2}\right)=\pi+\ln 2$

$4\arctan\!\left(\dfrac{y}{x}\right)+\ln\!\left(x^{2}+y^{2}\right)=\pi-\ln 2$

$4\arctan\!\left(\dfrac{y}{x}\right)-\ln\!\left(x^{2}+y^{2}\right)=-\pi+\ln 2$

$y=\dfrac{2}{x^{3}-3x}$

$y=\dfrac{2}{3x+x^{3}}$

$y=\dfrac{2}{x(3+x^{2})}$

$y=\dfrac{2}{x(3-x^{2})}$

General: $y=cx+\ln c\;(c>0)$. Singular: $y=-1+\ln\!\big(-\dfrac{1}{x}\big)$ valid for $x>0$

General: $y=cx+\ln c\;(c>0)$. Singular: $y=-1+\ln\!\big(-\dfrac{1}{x}\big)$ valid for $x<0$

General: $y=cx+\ln c\;(c>0)$. No real singular solution exists because $\dfrac{d}{dc}(cx+\ln c)=x+\dfrac{1}{c}$ never vanishes

General: $y=cx+\ln c\;(c>0)$. Singular: $y=-1+\ln\!\big(\dfrac{1}{x}\big)$ valid for $x>0$

The IVP has a unique solution $y(x)\equiv 0$ for all $x$

The IVP has exactly two solutions: $y\equiv 0$ and $y(x)=\left(\dfrac{2x}{3}\right)^{3/2}$ (suitably defined)

The IVP has infinitely many $C^{1}$ solutions; for any $a\ge 0$ the function $y_{a}(x)=0$ for $x\le a$ and $y_{a}(x)=\left(\dfrac{2}{3}(x-a)\right)^{3/2}$ for $x\ge a$ is a solution passing through $(0,0)$

No solution exists because the right-hand side is not Lipschitz at $y=0$