Joanna Benson

2021-12-28

Find the general/particular solution of the following homogeneous linear differential equations with constant coefficients.

$({D}^{3}+{D}^{2}-21D-45)y=0$

Daniel Cormack

Beginner2021-12-29Added 34 answers

Given differential equation

$({D}^{3}+{D}^{2}-21D-45)y=0$

Auxillary equation of D.E is

${r}^{3}+{r}^{2}-21r-45=0$

${r}^{3}-5{r}^{2}+6{r}^{2}-30r+9r-45=0$

${r}^{2}(r-5)+6r(r-5)+9(r-5)=0$

$(r-5)({r}^{2}+6r+9)$

$(r-5)({r}^{2}+2\left(r\right)\left(3\right)+{\left(3\right)}^{2})$

$(r-5){(r+3)}^{2}=0$

$r-5=0\text{}{(r+3)}^{2}=0$

$r=5\text{}r+3=0,r+3=0$

$r=-3-3$

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

Auxillary equation of D.E is

general solution

puhnut1m

Beginner2021-12-30Added 33 answers

Step 1

Given differential equation is

$({D}^{3}+{D}^{2}-21D-45)y=0$

Step 2

Its auxilary equation is

$f\left(m\right)={m}^{3}+{m}^{2}-21m-45=0$

$\Rightarrow {m}^{3}+{m}^{2}-21m-45=0$ (1)

put$m=5$

$LHS={5}^{3}+{5}^{2}-21\times 5-45=125+25-105-45$

$=150-150=0=RHS$

$\therefore (m-5)$ is a factor of f(m)

${m}^{2}(m-5)+(m-5)+9(m-5)=0$

$\Rightarrow (m-5)({m}^{2}+6m+9)=0$

$\Rightarrow (m-5)(m+3)(m+3)=0$

Therefore$m=5,-3,-3$

$\therefore C.F={c}_{1}{e}^{5x}+{c}_{2}{e}^{-3x}+{c}_{3}x{e}^{-3x}$

Solution is

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

Given differential equation is

Step 2

Its auxilary equation is

put

Therefore

Solution is

Vasquez

Expert2022-01-09Added 669 answers

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