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:

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

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

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

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