Problem 1 (5 points). First, let us assume W is a subset of...

Problem 1 (5 points). First, let us assume W is a subset of $mathbb{R}^2$ consisting of vectors of the form $egin{bmatrix} x_1\x_2 end{bmatrix}$. In each of the following cases, determine whether W is a subspace of $mathbb{R}^2$. (a) $W = {x: x_1 = 2x_2}$. (b) $W = {x: x_1 ext{ and } x_2 ext{ are rational numbers}}$. (c) $W = {x: x_1 - x_2 = 2}$. Next, we assume W is a subset of $mathbb{R}^3$ consisting of vectors of the form $x = egin{bmatrix} x_1\x_2\x_3 end{bmatrix}$. In each of the two cases, determine whether W is a subspace of $mathbb{R}^3$. (d) $W = {x: x_3 = x_1 + x_2}$. (e) $W = {x: x_3 = x_2 = 2x_1}$.

Transcript

00:01 For the solution of first equation where the equation is given as y double dash plus 5 pi dash plus 4 y which is equal to 10 to the power of minus 3 x.

00:14 So writing this in the operative form, its operator form will be as d squared plus 5 d plus 4 within the bracket and multiplied by pi which is equal to 10 e to the power of minus 6.

00:34 So the auxiliary equation, the auxiliary equation is f of m is equal to 0.

00:48 So now n squared plus 5m plus 4 is equal to 0.

00:54 So when we solve this quadratic equation by middle term splitting method, we will get our m plus 4 and m plus 1 as our 2 factors.

01:08 And therefore m will be equal to minus 1 or minus 4.

01:12 We have these two values of m here.

01:16 So now therefore the complementary function is y complementary function will be as c 1e to the power of minus x plus c2, e to the power of minus 4x.

01:40 So therefore y of p will be 1 .5 .5 will be 1 .1 .5.

01:44 Upon d plus 4 and d plus 1 multiplied by 10 to the bar of e i'm sorry 10 multiplied by e to the bar of minus 3 x further when we put down here values this is 1 upon minus 3 plus 4 minus 3 plus 1 multiplied by 10 multiplied by 2 2 2 2 2 2 2 2 2 2 2 2 2 2 x and that is equal to minus 5 e 2 2 2 2 2 x and that is equal to minus 5 to the bar of minus 3x.

02:18 So the above equation which we have in the question we can see here.

02:25 The rule used above is the prescient equation of identity.

02:28 We have f of r y is equal to e to the part of a x and therefore for this question as 1 by f of t, e to the power of a x will be equal to 1 upon f of l e to the power of a x.

02:49 If f of a does not equal to zero.

02:55 So now from this explanation, we can see the complete solution is y will be equal to yc plus y, and therefore y is equal to c1, e to the power of minus x plus c2, e to the part of minus 4x minus 5, e to the power of minus 3x.

03:23 So here we have our solution for first equation.

03:28 This is our answer for first part.

03:32 Now let's move on to the second part.

03:34 So in second equation, we have given here 10 y dash plus 50 y dash plus 57 into 6 .5...

Question

Please give Ace some feedback