Is my answer correct? if not please correct me

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

Which expression is equivalent to 5/6 ÷ 4

Ronch [10]

Answer:

5/24 or in decimal form 0.2083

Hope this helped

4 0

3 years ago

PLEASE HELP!!!!! What is the value of x in the matrix equation below?

Brums [2.3K]

The given matrix equation is,

$1.5\left[\begin{array}{cc}x&6\\8&4\end{array}\right] +y\left[\begin{array}{cc}1&4\\3&2\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right]$ .

Multiplying the matrices with the scalars, the given equation becomes,

$\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\$

Adding the matrices,

$\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\$

Matrix equality gives,

$1.5x+y=z\\ 9+4y=z\\ 12+3y=6z\\ 6+2y=2$

Solving the equations together,

$y=-2\\ 3y-1.5x=9\\ -1.5x=9+6\\ x=-10$

We can see that the equations are not consistent.

There is no solution.

3 0

2 years ago

Robert was using a bucket that holds 243.5 ounces (oz) of water when full. The bucket loses 0.3 oz of water per second. After 20

Assoli18 [71]

Answer:

237.5

Step-by-step explanation:

Since the bucket loses 0.3 oz of water per second, and 20 secs passed, you would have to multiply 0.3 and 20 to find out how much water leaked. And 0.3x20=6, so then, we would have to subtract 234.5 and 6 to find out how much water there is left in the bucket. 234.5-6=237.5

So 237.5 is your answer

4 0

2 years ago

Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown. What is the smallest poss

julia-pushkina [17]

Answer:

17 inches

Step-by-step explanation:

An obtuse triangle is the triangle in which one of the side is the longest. It contains an obtuse angle and the longest side is the side that is opposite to the vertex of the obtuse angle.

Let the three sides of the obtuse triangle be a, b and c respectively with c as the longest side. Let a = 9 inches and b = 14 inches.

Now we know that for an obtuse triangle,

$c^2 > a^2 +b^2$