DIFFERENTIAL EQUATIONS (MMR w3)

Which of the following is a separable differential equation?

• $\frac{\text{d}u}{\text{d}t}+\frac{2u}{t}=8$

• $\frac{\text{d}y}{\text{d}x}-6x^2y=9x^2$

• $xy'-2y-x^3\cos x=0$

• $\frac{d\theta}{dt}+2\theta=\sin\ t$

Solve the differential equation

 $\frac{dy}{dx}+2=2y$  .

•  $y=1+\sqrt{\frac{4}{-2x-A}}$  

•  $y=\frac{\beta e^x+2}{2}$  

•  $y=1+\alpha e^{2x}$  

•  $y=1+\frac{1}{-4x-A}$  

Find the general solution of the differential equation

 $2y\ dy-\left(y^2+1\right)\cos x\ dx=0$  

•  $y=\sqrt{Ae^{\left(\sin x-1\right)}}$  

•  $y=\sqrt{Ae^{\sin x}-1}$  

•  $y=\sqrt{\frac{A}{e^{\sin x}}-1}$  

•  $y=\sqrt{\sin x-1}$  

If

 $y=\frac{e^{\left(\alpha x+\beta\right)}}{\gamma}$  is the solution for differential equation  $\frac{dy}{dx}=6y$  , find the values of  $\alpha,$    $\beta$  and $\gamma$  when  $y=1,\ x=2$   .

•  $\alpha=1,\ \beta=0,\ \gamma=6$  

•  $\alpha=6,\ \beta=12,\ \gamma=6$  

•  $\alpha=12,\ \beta=6,\ \gamma=6$  

•  $\alpha=6,\ \ \beta=-12,\ \gamma=1$  

 If the gradient of a curve is given as

 $\frac{1}{x^2y}$  , find the equation of the curve at point  $\left(2,3\right)$  .

•  $y=\sqrt{\frac{10x-2}{x}}$  

•  $y=\sqrt{9-\ln16-\frac{2}{x}}$  

•  $y=\sqrt{-\frac{1}{x}+\frac{19}{2}}$  

Which of the following differential equation CAN NOT be separated?

• $\frac{dy}{dx}+1=4x$

• $y^2\frac{dy}{dx}=\sqrt{xy}-2x^2\sqrt{y}$

• $\frac{dy}{dx}+\frac{2}{x}y=\frac{1}{x+1}$

• $\frac{dy}{dx}-e^xy=e^x$

Do you think Week 3 questions are hard?

• Yes

• No

Given the curve $Ax^2\ +\ y^2=B$  is the solution of the differential equations  $\frac{dy}{dx}=-\ \frac{x}{y}$ , at   $\left(2,0\right)$  . Find the values of  $A$  and  $B.$  

•  $A=\frac{1}{2},\ B=\frac{1}{4}$  

•  $A=\frac{1}{2},\ B=1$  

•  $A=1,\ B=4$  

•  $A=-1,\ B=4$  

Find the general solution for the differential equation $\frac{dy}{dx}=xy$  .

•  $y=e^{\frac{x^2}{2}}+c$  

•  $y=\alpha e^{\frac{x^2}{2}}$  

•  $y=Ae^{x^2}$  

•  $y=\sqrt{\frac{1}{x^2+A}}$  

Given that the acceleration of an object satisfies the differential equation $e^t\frac{dv}{dt}=2\sqrt{v}$ . Find the equation of the velocity of the object if it started at  $v=1.$  

•  $v=\left(2-e^{-t}\right)^2$  

•  $v=\left(3-2e^{-t}\right)^2$  

•  $v=\left(2e^{-t}-1\right)^2$