Find the general solution.

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Dutsky Kamdon's picture
Find the general solution.

1. (D^5 + D^4 − 7D^3 − 11D^2 − 8D − 12)y = 0.
2. (D^4 − 2D^3 + 2D^2 − 2D + 1)y = 0.

8eightI'sD's picture
To get the general solution

To get the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$

We need to get the auxillary equation of the given differential equation above...

The auxillary equation would be:

$$r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$$

We need to get the roots of the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ using the rational zero theorem...

The rational zero theorem states that if $P(x)$ is a polynomial with integer coefficients and if is a zero of $P(x) ( P(\frac{p}{q} ) = 0 )$, then $p$ is a factor of the constant term of $P(x)$ and $q$ is a factor of the leading coefficient of $P(x)$.

With that in mind....

The factors of $p$ are:

$$-1,12,1,-12$$ $$-2,6,2,-6$$ $$-3,4,3,-4$$

The factors of $q$ are:

$$1,1,-1,-1$$

Then the values of $\frac{p}{q}$ would be:

$$-1,1,2,-3,4,-4,12,-12,6,-6,-2,3$$

It was found out that only $r= -2$ and $r= 3$ could make the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ true, instead of $5$

So, the factors of $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 $ would be $(r+2)(r-3)(unknown \space factor)$. To finally get the unknown factor in $(r+2)(r-3)(unknown \space factor)$, we need to divide $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12$ by $(r+2)(r-3)$ or $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12$ divided by $r^2-r-6$.

Turns out that the quotient of $\frac{r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12}{(r+2)(r-3)}$ is the unknown factor $r^3 + 2r^2 + r + 2$

So, the factors of $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 $ would be $(r+2)(r-3)(r^3 + 2r^2 + r + 2) $

To get the roots of the equation $r^3 + 2r^2 + r + 2 = 0$, use again the rational zero theorem.

With that in mind...

The factors of $p$ are:
$$2,1,-2,-1$$

The factors of $q$ are:

$$,1,1-1,-1$$

Then the values of $\frac{p}{q}$ would be:

$$2,1,-2,-1$$

It was found out that only $r= -2$ could make the polynomial $r^3 + 2r^2 + r + 2 = 0$ true, instead of $3$ roots.

So...the factors of $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 $ would now be $(r+2)(r-3)[(r+2)(unknown \space factor)] $

To finally get the unknown factor in $(r+2)(r-3)[(r+2)(unknown \space factor)] $, we need to divide $r^3 + 2r^2 + r + 2$ by $r+2$ It turns out that the quotient of $\frac{r^3 + 2r^2 + r + 2}{r+2}$ is $r^2+1$.

So, the factors of $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 $ would be $(r+2)(r-3)(r+2)(r^2+1)$

Getting the roots of the equation $r^2+1 = 0$, the roots would be $r=-i$ and $r = +i$

After a long and tedious process, the factors of $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12$ would be $(r+2)(r-3)(r+2)(r+i)(r-i)$. The roots of the auxillary equation $r^5 + r^4 -7r^3 - 11r^2 - 8r - 12 = 0$ or $(r+2)(r-3)(r+2)(r+i)(r-i) = 0$ or $(r+2)(r+2)(r-3)(r+i)(r-i) = 0$ is $r = -2,-2,3,-i, +i$

The general solution to the given differential equation is:

$$y = c_1e^{-2x} + c_2xe^{-2} + c_3e^{3x} + c_4\cos x + c_5\sin x$$

Let's break it down.

The first two factors of $(r+2)(r+2)(r-3)(r+i)(r-i) = 0$, which are $(r+2)(r+2)$, have repeating roots, so the general solution that portion would be $y = c_1e^{rx}+c_2xe^{rx}$
or $y = c_1e^{-2x}+c_2xe^{-2x}$

The third factor, which is the $r = 3$, is nothing special, so the general solution that portion would be $y = c_3e^{rx}$ or $y = c_3e^{3x}$

The last two factors, which are $(r+i)(r-i)$ have imaginary roots written in the complex form $a+bi$, which is $r = 0+i$ and $r = 0-i$, so the general solution that portion would be $y = e^{ax}(c_4\cos (bx)+c_5\sin (bx))$ or $y = e^{0x}(c_4\cos x+c_5 \sin x)$ or $y = (c_4\cos x+c_5 \sin x)$

Alternate ways of solving are highly recommended...

Sa ikalawang tanong....halos parehas lang ng solution....medyo mahaba rin....pero makukuha nyo po yan....hihiihi

8eightI'sD's picture
Merry Christmas po mga "Math"

Merry Christmas po mga "Math"-alinong Pinoys:-D