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

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 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 obtained as $y=\large\frac{1}{f(D)}$$\times h$
  • The GS is $y=CF+PI$
  • Let $m_1,m_2$ be the roots of the CE
  • Case 1: $m_1,m_2$ are real numbers and distinct
  • $CF=Ae^{m_1x}+Be^{m_2x}$
  • Case 2: $m_1,m_2$ are complex (i.e)$m_1=\alpha+i\beta$ and $m_2=\alpha-i\beta$
  • Then $CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x]$
  • Case 3: $m_1,m_2$ are real and equal say $m_1$
  • $CF=(A+Bx)e^{m_1x}$
  • Take $y=C_0+C_1x$ for $X=x$
  • and $y=C_0+C_1x+C_2x^2$ for $X=x^2$
  • $\therefore (aD^2+bD+c)y=f(x)$
  • By substituting for $y$ and comparing coeffs,$C_0,C_1$ and $C_2$ can be determined.
Step 1:
CE : $m^2+3m-4=0$
CF : $Ae^{-4x}+Be^x$
Step 2:
Let the $PI=C_0+C_1+C_2x^2$
$\therefore (D^2+3D-4)(C_0+C_1x+C_2x^2)=x^2$
Step 3:
Comparing coeff on LHS and RHS
Constant $\Rightarrow -4C_0+3C_1+2C_2=0$---------(1)
$x$ term $\Rightarrow -4C_1+6C_2=0$-------(2)
$x^2$ term $\Rightarrow -4C_2=1$
From equ(2)
Substituting $C_1=-\large\frac{3}{8},$$C_2=-\large\frac{1}{4}$ in equ(1)
Step 4:
GS :$y=Ae^{-4x}+Be^x-\large\frac{13}{32}-\frac{3}{8}$$x-\large\frac{1}{4}$$x^2$
answered Sep 6, 2013 by sreemathi.v