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Solve the following differential equation;$(D^{2}-4D+13)$y$=e^{-3x}$

1 Answer

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  • 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:
Characterstic equation :$m^2-4m+13=0$
$m=\large\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm 6i}{2}$$=2\pm 3i$
$CF =e^{2x}[A\cos 3x+B\sin 3x]$
Step 2:
Step 3:
GS : $y=Ae^{2x}[A\cos 3x+B\sin 3x]+\large\frac{e^{-3x}}{34}$
answered Sep 5, 2013 by sreemathi.v
edited Sep 8, 2013 by sreemathi.v