. During the Grinch's engine test, if the path traveled is represented by the

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

4 0

Answer: $f(-2)=-6$

Step-by-step explanation:

For this exercise it is important to remember the multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-$

You have the following function provided in the exercise, which represents the path traveled:

$f(x) = 4x + 2$

In order to find the value of $f(-2)$ you can follow these steps:

1. You need to substitute $x=-2$ into the function, then:

$f(-2) = 4(-2)+ 2$

2. Finally you must evaluate. Then you get:

$f(-2) =-8+ 2\\\\f(-2)=-6$

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Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$