logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b $y=e^{2x}(a+bx)$

$\begin{array}{1 1} (A)\;y'' = 4be^{2x} + 4e^{2x}(a+bx) \\ (B)\;y'' = 4be^{x} + 4e^{2x}(a+bx) \\ (B)\;y'' = 4be^{x} + 4e^{2x}(a+bx) \\(D)\;y'' = 4be^{2x} + 4e^{x}(a+bx)\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • When the function occurs in the form of their products, we can differentiate them using product rule, which states that $uv = uv' + vu' $
Step 1:
Given: $y = e^{2x}(a + bx)$
Differentiating on both sides by applying the product rule
let $u$ be $e^{2x}$ hence $u'$ will be $2e^2x$ and $v = a+bx$, hence $v' = b$
Hence $y' = 2e^{2x}(a+bx) + be^{2x}$
Step 2:
substituting $y = e^{2x}(ax+b)$
$y' = 2y + be^{2x}$
$y' - 2y = be^{2x}$-----(1)
Differentiating $y'$ again by applying the product rule we get
$y'' = 2be^{2x} + 4e^{2x}(a+bx) + 2e^{2x}(b)$
$y'' = 4be^{2x} + 4e^{2x}(a+bx)$
answered Aug 16, 2013 by sreemathi.v
 

Related questions

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
...