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Solve the following differential equation; $(3D^{2}+4D+1)$$y=3e^{-\Large\frac{x}{3}}$

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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}}$
answered Sep 6, 2013 by sreemathi.v
 
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