# Solve the following differential equation;$\large\frac{d^{2}y}{dx^{2}}$$-3\large\frac{dy}{dx}$$+2y=2e^{3x}$ when $x =\log$$2,y=0, and when x=0,y=0 ## 1 Answer Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X • Where X is a function of x • The solution is obtained in two parts. • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI • The GS is y=CF+PI • Let m_1,m_2 be the roots of the CE • Case 1: m_1,m_2 are real numbers and distinct • CF=Ae^{m_1x}+Be^{m_2x} • Case 2: m_1,m_2 are complex (i.e)m_1=\alpha+i\beta and m_2=\alpha-i\beta • Then CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x] • Case 3: m_1,m_2 are real and equal say m_1 • CF=(A+Bx)e^{m_1x} • The PI when X=e^{\alpha x},\alpha is a constant • Case 1: f(\alpha)\neq 0 • PI=\large\frac{1}{f(D)}e^{\alpha x}=\frac{e^{\alpha x}}{f(\alpha)} • Case 2: f(\alpha)\neq 0(\alpha=m_1, one of the roots of the CE) • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{1}{(D-\alpha)\theta(D)}e^{\alpha x} where \theta(\alpha)\neq 0 • \Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)} • Case 3: f(\alpha)=0 and m_1=m_2=\alpha then • \large\frac{1}{f(D)}e^{\alpha x}=\large\frac{x^2}{2}e^{\alpha x} Step 1: CE :m^2-3m+2=0 (m-1)(m-2)=0 m_1=1,m_2=2 CF :Ae^x+Be^{2x} Step 2: PI=\large\frac{1}{D^2-3D+2}$$2e^{3x}\;\;(\alpha=3\neq m_1$ or $m_2)$
$\;\;=\large\frac{2e^{3x}}{9-9+2}=e^{3x}$
Step 3:
GS: $y=Ae^x+Be^{2x}+e^{3x}$
Step 4:
When $x=\log 2,y=0$
$0=A+Ae^{\log 2}+Be^{2\log 2}+e^{3\log 2}$
$0=2A+4B+8$-------(1)
When $x=0,y=0$
$0=A+B+1$-------(2)
Solving : $2A+4B=-8$
$\qquad\;\;\; 2A+2B=-2$
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$2B=-6\Rightarrow B=-3$
$\therefore A=-1-B=-1+3=2$
The required solution is
$y=2e^x-3e^{2x}+e^{3x}$
$\;\;\;=e^x[2-3e^x+e^{2x}]$