Solve the following differential equation; $(D^{2}+3D-4)$$y=x^{2} 1 Answer Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X • Where X is a function of x • The solution is obtained in two parts. • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI obtained as y=\large\frac{1}{f(D)}$$\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 : $m^2+3m-4=0$
$(m+4)(m-1)=0$
$m_1=-4,m_2=1$
CF : $Ae^{-4x}+Be^x$
Step 2:
Let the $PI=C_0+C_1+C_2x^2$
$\therefore (D^2+3D-4)(C_0+C_1x+C_2x^2)=x^2$
$-4C_0+3C_1-4C_1x+2C_2+6C_2x-4C_2x^2=x^2$
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 $\Rightarrow -4C_2=1$
$C_2=\large\frac{-1}{4}$------(3)
From equ(2)
$4C_1=6C_2=\large\frac{-6}{4}$
$C_1=-\large\frac{-6}{16}=\frac{-3}{8}$
Substituting $C_1=-\large\frac{3}{8},$$C_2=-\large\frac{1}{4} in equ(1) -4C_0-\large\frac{9}{8}-\frac{1}{2}$$=0$
$4C_0=\large\frac{-13}{8}$
$C_0=\large\frac{-13}{32}$
Step 4:
GS :$y=Ae^{-4x}+Be^x-\large\frac{13}{32}-\frac{3}{8}$$x-\large\frac{1}{4}$$x^2$