Question

1. Find the general solution of the given higher-order differential equation.

d^3u/ dt^3 + d^2u/ dt^2 − 2u = 0


u(t) = ____


2. Find the general solution of the given higher-order differential equation.

y''' − 9y'' + 15y' + 25y = 0


y(x) =____

Answer 1

$u^{\prime \prime \prime}+ u^{\prime \prime}-2u=0 \\[1 em] \text { A\,linear homogeneous\, equation }\\[1 em] \text { The general solution \,is}\\[1 em] u(t)=u_{h} \\[1 em] r^{3}+ r^{2}-2=0 \implies p=\pm 2 ,\pm1 ,q=\pm1 \\[1 em] \frac{p}{q}=\pm 1 ,\pm 2 ,.....\\[1 em] f(1)=0 \to ~ 1~ \text{ is a root of the equation} \\[1 em] \text{ Divide} ~r^{3}+ r^{2}-2 ~\text{ by} ~r-1 ~\text{ we get }r^2+2r+2 \\[1 em] \text{Use quadratic Formula }\rightarrow r=-1+i ,r=-1-i ,r=1 \\[1 em] \therefore u(t)=c_{1}e^{ t} +e^{- t}(c_{2} \cos t +c_{3} \sin t) ~~~~~(1) \\[3 em] y^{\prime \prime \prime}-9y^{\prime \prime}+15y^{ \prime}+25y=0 \\[1 em] \text { A\,linear homogeneous\, equation }\\[1 em] \text { The general solution \, is }\\[1 em] y(x)=y_{h} \\[1 em] r^{3}-9r^{2}+15r+25=0 \implies p=\pm 5 ,\pm1 , \pm25 ,q=\pm1 \\[1 em] \frac{p}{q}=\pm 1 ,\pm 5 ,....\\[1 em] f(-1)=0 \to ~ 1~ \text{ is a root of the equation} \\[1 em] \text{ Divide} ~r^{3}-9r^{2}+15r+25 ~\text{ by} ~r+1 ~\text{ we get }r^2-10r+25 \\[1 em] r^2-10r+25 =(r-5)(r-5)\rightarrow r=5 ,r=5 ,r=-1 \\[1 em] \therefore y(x)=c_{1}e^{ -x} +c_{2} e^{ 5x} +c_{3}x e^{ 5x} ~~~~~(2) \\[1 em]$