# Solve the following differential equation;$(D^{2}-6D+9)$$y=x+e^{2x} ## 1 Answer Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=x • Where X is a function of x • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI • The GS is y=CF+PI • Let m_1,m_2 be the roots of the CE • Case 1: m_1,m_2 are real numbers and distinct • CF=Ae^{m_1x}+Be^{m_2x} • Case 2: m_1,m_2 are complex (i.e)m_1=\alpha+i\beta and m_2=\alpha-i\beta • Then CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x] • Case 3: m_1,m_2 are real and equal say m_1 • CF=(A+Bx)e^{m_1x} • The PI when X=e^{\alpha x},\alpha is a constant • Case 1: f(\alpha)\neq 0 • PI=\large\frac{1}{f(D)}e^{\alpha x}=\frac{e^{\alpha x}}{f(\alpha)} • Case 2: f(\alpha)\neq 0(\alpha=m_1, one of the roots of the CE) • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{1}{(D-\alpha)\theta(D)}e^{\alpha x} where \theta(\alpha)\neq 0 • \Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)} • Case 3: f(\alpha)=0 and m_1=m_2=\alpha then • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{x^2}{2}e^{\alpha x} • Take y=C_0+C_1x for X=x • and y=C_0+C_1x+C_2x^2 for X=x^2 • \therefore (aD^2+bD+c)y=f(x) • By substituting for y and comparing coeffs,C_0,C_1 and C_2 can be determined. Step 1: CE :m^2-6m+9=0 (m-3)^2=0 m_1=m_2=3 CF :e^{3x}(Ax+B) Step 2: PI =\large\frac{1}{D^2-6D+9}$$x+\large\frac{1}{D^2-6D+9}$$e^{2x} Step 3: PI_1=\large\frac{1}{D^2-6D+9}$$x$
Let $PI_1=C_0+C_1x$
$\therefore(D^2-6D+9)(C_0+C_1x)=x$
Step 4:
$9C_0+9C_1x-6C_1=x$
Comparing coeffs
$x$ term :$C_1=\large\frac{1}{9}$
Constant : $9C_0-\large\frac{6}{9}$$=0 \Rightarrow C_0=\large\frac{2}{27} PI_1=\large\frac{2}{3}$$+x$
Step 5:
$PI_2=\large\frac{1}{D^2-6D+9}$$e^{2x}=\large\frac{e^{2x}}{4-12+9}$$=e^{2x}$
Step 6:

1 answer

1 answer

1 answer