FIRST ORDER DIFFERENTIAL EQUATIONS

What is the integrating factor for the differential equation:  $\frac{1}{x}\frac{\text{d}x}{\text{d}y}-\frac{1}{1+x^2}y=x^3$  

•  $\ln\left(x^2\right)$  

•  $-\frac{1}{x^2}$  

•  $-\frac{1}{\sqrt{x}}$  

•  $\frac{1}{\sqrt{x}}$  

Which of the following statement is TRUE.

• $\frac{\text{d}x}{\text{d}y}-\frac{2}{x}=e^x$ 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.

 $\frac{\text{d}x}{\text{d}y}-4y-5+e^{-2x}=0$  

The general solution for the above DE is,

•  $y=-20+6e^{-2x}+Ce^{4x}$  

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

•  $y=-20e^{-8x}+e^{-10x}+Ce^{-4x}$  

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

Integrating factor of the differential equation  $x\frac{\text{d}y}{\text{d}x}-y=\sin x$  

•  $-\frac{1}{x}$  

•  $\frac{1}{x}$  

•  $\ln\left|x\right|$  

•  $\frac{1}{x^2}$  

The solution of linear differential equation  $x\frac{\text{d}y}{\text{d}x}+2y=x^2$  

•  $y=\frac{x^2+c}{4x^2}$  

•  $y=\frac{x^2}{4}+c$  

•  $y=\frac{x^4+c}{x^2}$  

•  $y=\frac{x^4+c}{4x^2}$  

Which of the following mathematical models CANNOT be solved by using separation of variables

• $\frac{\text{d}P}{\text{dt}}=kP$

• $\frac{dT}{dt}=k\left(T-Tm\right)$

• $\frac{dA}{dt}=kA$

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

By separation of variables, solve the resulting equations $\int\ \frac{v}{v+1}dv=\int_{ }^{ }\ \frac{1}{y}dy\text{}$  

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

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

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

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

 Solve the given differential equations by separable variable method   $\frac{\text{d}v}{\text{d}t}=\frac{3+v^2}{v}$  

•  $v=\ln\left|3+v^2\right|+C$  

•  $v^2=e^{2t+2c}-3$  

•  $v=Ae^{2t}+3,\ \ \ \ \ A=e^{2c}$  

•  $v^2=Ae^{2t}+3,\ \ \ \ \ \ A=C$  

 Solve the given differential equations by separable variable method   $\frac{\text{d}y}{\text{d}x}=e^{-y}\left(2x-4\right),\ \ \ \ \ y\left(5\right)=0$  

•  $y=\ln\left|x^2-4x-4\right|$  

•  $y=x^2-4x-4$  

•  $y=4e^{2t}+4\ \ \ \ \ A=e^{2c}$  

•  $y^2=4e^{2x}+2,\ \ \ \ \ \ A=C$  

The solution of linear differential equation  $\frac{\text{d}y}{\text{d}x}+y=e^x$  

•  $y=\frac{e^2+c}{4x^2}$  

•  $y=\frac{ex^2}{4}+c$  

•  $y=\frac{e^{4x}+c}{x^2}$  

•  $y=\frac{e^x}{2}+Ce^{-x}$