Write a linear equation in the form of y=mx+b

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

4 years ago

(6x² + 36x² + 12) ÷ (x+6)

astra-53 [7]

Answer:

=42x2+1

Step-by-step explanation:

6x2+36x2+12x+6

=42x2+1

8 0

3 years ago

All geometric figures,shapes,and solids can be named using sets of

Naddika [18.5K]

Answer:

All geometric figures, shapes and solids can be named as a sets of surface regions and plane regions and they all lie in three dimensional spaces.

Step-by-step explanation:

7 0

3 years ago

50.441−6.203=? help asap

Doss [256]

Answer:

44.238

Step-by-step explanation:

50.441-6.203=44.238

6 0

3 years ago

Read 2 more answers