Ask Questions, Get Answers


Solve the following differential equation;$(D^{2}-6D+9)$$y=x+e^{2x}$

1 Answer

  • 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 :$m^2-6m+9=0$
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:
Let $PI_1=C_0+C_1x$
Step 4:
Comparing coeffs
$x$ term :$C_1=\large\frac{1}{9}$
Constant : $9C_0-\large\frac{6}{9}$$=0$
$\Rightarrow C_0=\large\frac{2}{27}$
Step 5:
Step 6:
GS : $y=Ae^{3x}(Ax+B)+\large\frac{2}{27}+\frac{x}{9}$$+e^{2x}$
answered Sep 6, 2013 by sreemathi.v