$\frac{\text{d}x}{\text{d}y}-\frac{2}{x}=e^x$ is linear and its integrating factor is $-\frac{1}{x^2}.$

$\frac{\text{d}x}{\text{d}y}-4x=0\$ with condition $y\left(1\right)=2$ is linear and its solution is $y=2x^2.$

$\frac{\text{d}x}{\text{d}y}-\frac{2}{x}=e^x\$ is linear and its solution is $y=\frac{1}{x^2}.$

$y^3dx-4x^2dy=0$ is not exact.

 $y=-20+6e^{-2x}+Ce^{4x}$  

 $y=-\frac{5}{4}e^{-4x}+\frac{1}{6}e^{-2x}+Ce^{4x}$  

 $y=-20e^{-8x}+e^{-10x}+Ce^{-4x}$  

 $y=-\frac{5}{4}+\frac{1}{6}e^{-2x}+Ce^{-4x}$  

 $-\frac{1}{x}$  

 $\frac{1}{x}$  

 $\ln\left|x\right|$  

 $\frac{1}{x^2}$  

 $y=\frac{x^2+c}{4x^2}$  

 $y=\frac{x^2}{4}+c$  

 $y=\frac{x^4+c}{x^2}$  

 $y=\frac{x^4+c}{4x^2}$  

$\frac{\text{d}P}{\text{dt}}=kP$

$\frac{dT}{dt}=k\left(T-Tm\right)$

$\frac{dA}{dt}=kA$

$L\ \frac{dI}{dt}+RI=E\left(t\right)$

 $\frac{x}{y}-\ln\left|\frac{x}{y}+1\right|=\ln y+C$  

 $\frac{x}{y}+\ln\left|\frac{x}{y}+1\right|=\ln y+C$  

 $\frac{x}{y}+\frac{x^2}{2y^2}=\ln y+C$  

 $\frac{x}{y}-\frac{x^2}{2y^2}=\ln y+C$  

 $v=\ln\left|3+v^2\right|+C$  

 $v^2=e^{2t+2c}-3$  

 $v=Ae^{2t}+3,\ \ \ \ \ A=e^{2c}$  

 $v^2=Ae^{2t}+3,\ \ \ \ \ \ A=C$  

 $y=\ln\left|x^2-4x-4\right|$  

 $y=x^2-4x-4$  

 $y=4e^{2t}+4\ \ \ \ \ A=e^{2c}$  

 $y^2=4e^{2x}+2,\ \ \ \ \ \ A=C$  

 $y=\frac{e^2+c}{4x^2}$  

 $y=\frac{ex^2}{4}+c$  

 $y=\frac{e^{4x}+c}{x^2}$  

 $y=\frac{e^x}{2}+Ce^{-x}$