logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Solve the following differential equation;$\large\frac{d^{2}y}{dx^{2}}$$-3\large\frac{dy}{dx}$$+2y=2e^{3x}$ when $x =\log$$2$,$y=0,$ and when $x=0,y=0$

Can you answer this question?
 
 

1 Answer

0 votes
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 :$m^2-3m+2=0$
$(m-1)(m-2)=0$
$m_1=1,m_2=2$
CF :$Ae^x+Be^{2x}$
Step 2:
PI=$\large\frac{1}{D^2-3D+2}$$2e^{3x}\;\;(\alpha=3\neq m_1$ or $m_2)$
$\;\;=\large\frac{2e^{3x}}{9-9+2}=e^{3x}$
Step 3:
GS: $y=Ae^x+Be^{2x}+e^{3x}$
Step 4:
When $x=\log 2,y=0$
$0=A+Ae^{\log 2}+Be^{2\log 2}+e^{3\log 2}$
$0=2A+4B+8$-------(1)
When $x=0,y=0$
$0=A+B+1$-------(2)
Solving : $2A+4B=-8$
$\qquad\;\;\; 2A+2B=-2$
___________________________
$2B=-6\Rightarrow B=-3$
$\therefore A=-1-B=-1+3=2$
The required solution is
$y=2e^x-3e^{2x}+e^{3x}$
$\;\;\;=e^x[2-3e^x+e^{2x}]$
answered Sep 6, 2013 by sreemathi.v
edited Sep 8, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...