Joanna Benson

2021-12-28

Find the general/particular solution of the following homogeneous linear differential equations with constant coefficients.
$\left({D}^{3}+{D}^{2}-21D-45\right)y=0$

### Answer & Explanation

Daniel Cormack

Given differential equation
$\left({D}^{3}+{D}^{2}-21D-45\right)y=0$
Auxillary equation of D.E is
${r}^{3}+{r}^{2}-21r-45=0$
${r}^{3}-5{r}^{2}+6{r}^{2}-30r+9r-45=0$
${r}^{2}\left(r-5\right)+6r\left(r-5\right)+9\left(r-5\right)=0$
$\left(r-5\right)\left({r}^{2}+6r+9\right)$
$\left(r-5\right)\left({r}^{2}+2\left(r\right)\left(3\right)+{\left(3\right)}^{2}\right)$
$\left(r-5\right){\left(r+3\right)}^{2}=0$

$r=-3-3$
general solution $y={c}_{1}{e}^{5x}+\left({c}_{2}+{c}_{3}x\right){e}^{-3x}$

puhnut1m

Step 1
Given differential equation is
$\left({D}^{3}+{D}^{2}-21D-45\right)y=0$
Step 2
Its auxilary equation is
$f\left(m\right)={m}^{3}+{m}^{2}-21m-45=0$
$⇒{m}^{3}+{m}^{2}-21m-45=0$ (1)
put $m=5$
$LHS={5}^{3}+{5}^{2}-21×5-45=125+25-105-45$
$=150-150=0=RHS$
$\therefore \left(m-5\right)$ is a factor of f(m)
${m}^{2}\left(m-5\right)+\left(m-5\right)+9\left(m-5\right)=0$
$⇒\left(m-5\right)\left({m}^{2}+6m+9\right)=0$
$⇒\left(m-5\right)\left(m+3\right)\left(m+3\right)=0$
Therefore $m=5,-3,-3$
$\therefore C.F={c}_{1}{e}^{5x}+{c}_{2}{e}^{-3x}+{c}_{3}x{e}^{-3x}$
Solution is
$y\left(x\right)={c}_{1}{e}^{5x}+\left({c}_{2}+{c}_{3}x\right){e}^{-3x}$

Vasquez

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