Second Order Linear Differential Equations

What is the general solution to the DE $\frac{\text{d}^2x}{\text{d}y^2}+7\frac{\text{d}x}{\text{d}y}-8y=0$  ?

•  $y=Ce^{-x}+De^{8x}$  

•  $y=Ce^x+De^{-8x}$  

•  $y=Ce^x+De^{7x}$  

•  $y=Ce^{-x}+De^{7x}$  

 $\frac{\text{d}^2x}{\text{d}y^2}+6\frac{\text{d}x}{\text{d}y}-5y=0\$  

The general solution to the DE is,

•  $y=Ae^x+Be^{5x}$  

•  $y=Ae^{-x}+Be^{-5x}$  

•  $y=Ae^{\left(-3+\sqrt{14}\right)x}+Be^{\left(-3-\sqrt{14}\right)x}$  

•  $y=A\cos\left(-3+\sqrt{14}\right)x+B\sin\left(-3-\sqrt{14}\right)x$  

The roots of the auxiliary equation 
 $\frac{\text{d}x}{\text{d}y}+16y=0$  is

•  $y=A\cos16x+B\sin16x$  

•  $y=A+Be^{-16x}$  

•  $y=A\cos4x+B\sin4x$  

•  $y=Ae^{4x}+Be^{-4x}$  

The general solution of the DE
 $\frac{\text{d}^2x}{\text{d}y^2}+4\frac{\text{d}x}{\text{d}y}-5y=0$  is

•  $y=Ae^x+Be^{-5x}$  

•  $y=Ae^{-x}+Be^{5x}$  

•  $y=Ae^x+Be^{5x}$  

•  $y=Ae^{-x}+Be^{-5x}$  

 $\frac{\text{d}^2x}{\text{d}y^2}+4\frac{\text{d}x}{\text{d}y}-6y=0$  

Which of the following options are TRUE about the above DE?

• The roots of the auxiliary equations are two complex roots.

• The auxiliary equation has two different roots.

• The auxiliary equation has two equal roots.

• The equation is homogeneous.

The initial guess for $f\left(x\right)-4\cos3x+2x^2$  

•  $y_p=\left(C\cos3x+Fx^2\right)$  

•  $y_p=\left(C\cos3x+D\sin3x+Fx^2+Gx+H\right)$  

•  $y_p=\left(C\cos3x+D\sin3x+Fx^2\right)$  

•  $y_p=\left(C\cos3x-D\sin3x+Fx^2+G\right)$  

Given  $\frac{\text{d}^2x}{\text{d}y^2}+4\frac{\text{d}x}{\text{d}y}+3y=2e^{-x}.$  

The correct  $y_p$  is

•  $Ce^{-x}$  

•  $C\cos x+D\sin x$  

•  $Ce^x$  

•  $Cxe^{-x}$  

 $\frac{\text{d}^2x}{\text{d}y^2}-y=\cot x$  
The above DE can be solved using the undetermined coefficient meThod.

• TRUE

• FALSE

 $x^2\frac{\text{d}^2x}{\text{d}y^2}-5y=x+1$  is a non homogeneous second order differential equations with constant coefficients.

• TRUE

• FALSE

Do you understand Chapter 2 well? Are you ready for the Test next week? You can put comment here.