MATH34032_Video3_variationofparameters

Suppose that a complementary solution is

 $u_c\left(x\right)=c_1u_1\left(x\right)+c_2u_2\left(x\right).$  The ansatz for the particular solution in variation of parameters ... (Select all options which are correct.)

• ... is $u_p\left(x\right)=v_1\left(x\right)u_1\left(x\right)+v_2\left(x\right)u_2\left(x\right).$ 

• ... depends on the right hand side of the ODE.

• ... is the same as the complementary solution.

• ... only works in some cases.

Consider the ODE

 $u''=f\left(x\right)$ . Which of the following is a complementary solution?

•  $u_c\left(x\right)=c_1x+c_2.$  

• A second integral of  $f$ .

•  $u_c\left(x\right)=c_1x+c_2\ln\left(x\right)x$  

•  $u_c\left(x\right)=c_1+c_2\ln\left(x\right)$  

Answer explanation

For the next question we will use the complementary solution $u_c\left(x\right)=c_1x+c_2$  (i.e.  $u_1\left(x\right)=x$  and $u_2\left(x\right)=1$ )

Using the complementary solution from the last question, variation of paramters will give which of the following as a particular solution for the ODE

 $u''=f\left(x\right)?$ 

•  $u_p\left(x\right)=\int_{-\infty}^x\left(x-x_0\right)f\left(x_0\right)dx_0$  

•  $u_p\left(x\right)=\int_{-\infty}^x\left(x_0-1\right)f\left(x_0\right)dx_0$  

•  $u_p\left(x\right)=\int_{-\infty}^{x_0}\left(x-x_0\right)f\left(x_0\right)dx_0$  

•  $u_p\left(x\right)=\int_{-\infty}^x\left(x-1\right)f\left(x_0\right)dx_0$  

Using the same method as the example in the video, which of the following would be a Green's function for the differential operator

 $L=\frac{\text{d}^2}{\text{d}x^2}+4?$  

•  $G=\frac{1}{2}\sin\left(2\left(x-x_0\right)\right)H\left(x-x_0\right)$ 

•  $G=\frac{1}{4}\sin\left(4\left(x-x_0\right)\right)H\left(x-x_0\right)$ 

•  $G=\frac{1}{2}\cos\left(2\left(x-x_0\right)\right)H\left(x-x_0\right)$ 

•  $G=\frac{1}{4}\cos\left(4\left(x-x_0\right)\right)H\left(x-x_0\right)$ 

Use variation of parameters to find a particular solution for the ODE

 $x^2u''-6u=f\left(x\right)$  on the domain  $x>0$ .

•  $u_p\left(x\right)=\int_1^x\left(\frac{x^3}{x_0^2}-\frac{x_0^3}{x^2}\right)\frac{f\left(x_0\right)}{5x_0^2}^{ }dx_0$  

•  $u_p\left(x\right)=\int_1^x\left(\frac{x_0^3}{x_{ }^2}-\frac{x_{ }^3}{x_0^2}\right)\frac{f\left(x_0\right)}{5x_0^2}dx_0$  

•  $u_p\left(x\right)=\int_1^x\left(\frac{x^3}{x_0^2}-\frac{x_0^3}{x^2}\right)\frac{f\left(x_0\right)}{5x_{ }^2}dx_0$  

•  $u_p\left(x\right)=\int_1^{x_0}\left(\frac{x^3}{x_0^2}-\frac{x_0^3}{x^2}\right)\frac{f\left(x_0\right)}{5x_0^2}dx_0$  

Answer explanation

Note that since the domain is  $x>0$ , the lower limit in the integral should be positive.