# Solve the following differential equation; $(3D^{2}+4D+1)$$y=3e^{-\Large\frac{x}{3}} ## 1 Answer Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X • Where X is a function of x • The solution is obtained in two parts. • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI • The GS is y=CF+PI • Let m_1,m_2 be the roots of the CE • Case 1: m_1,m_2 are real numbers and distinct • CF=Ae^{m_1x}+Be^{m_2x} • Case 2: m_1,m_2 are complex (i.e)m_1=\alpha+i\beta and m_2=\alpha-i\beta • Then CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x] • Case 3: m_1,m_2 are real and equal say m_1 • CF=(A+Bx)e^{m_1x} • The PI when X=e^{\alpha x},\alpha is a constant • Case 1: f(\alpha)\neq 0 • PI=\large\frac{1}{f(D)}e^{\alpha x}=\frac{e^{\alpha x}}{f(\alpha)} • Case 2: f(\alpha)\neq 0(\alpha=m_1, one of the roots of the CE) • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{1}{(D-\alpha)\theta(D)}e^{\alpha x} where \theta(\alpha)\neq 0 • \Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)} • Case 3: f(\alpha)=0 and m_1=m_2=\alpha then • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{x^2}{2}e^{\alpha x} Step 1: CE : 3m^2+4m+1=0 3m^2+3m+m+1=0 3m(m+1)+1(m+1)=0 (3m+1)(m+1)=0 m_1=-\large\frac{-1}{3}$$,m_2=-1$
CF=$Ae^{-\large\frac{1}{3}}x+Be^{-x}$
Step 2:
PI =$\large\frac{1}{3D^2+4D+1}$$3e^{-\Large\frac{x}{3}} \;\;\;=\large\frac{3}{(3D+1)(D+1)}$$e^{-\Large\frac{x}{3}}$
$\;\;\;=\large\frac{3}{3(D+\large\frac{1}{3})(\large\frac{-1}{3}+1)}$$e^{-\Large\frac{x}{3}} \;\;\;=\large\frac{3}{2}\frac{1}{(D+\large\frac{1}{3})}e^{-\Large\frac{x}{3}} \;\;\;=\large\frac{3xe^{-\Large\frac{x}{3}}}{2} Step 3: GS : y=Ae^{\Large\frac{-x}{3}}+Be^{-x}+\large\frac{3}{2}$$xe^{\Large\frac{-x}{3}}$