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

Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is $y'=e^x\;\sin x.$

$\begin{array}{1 1} 2y - 1 = e^{\large x(\sin x - \cos x)} \\ 2y + 1 = e^{\large x(\sin x - \cos x)} \\ 2y - 1 = e^{\large x(\sin x + \cos x)} \\2y - 1 = e^{\large x(\cos x - \sin x)} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • When the function appears in the form of integration of $udv$ then it can be integrated by parts which is $uv - \int vdu$
Step 1:
Given $\large\frac{dy}{dx}$$ = e^x\sin x$
seperating the variables $dy = e^x\sin x$
$\int dy = \int e^x\sin x dx$
$\int e^{x\sin x} dx$ can be done by the method of integration by parts.
Let $u = e^x$; hence $du = e^xd$x and $dv = \sin xdx$; so $v =- \cos x$
Step 2:
$ uv - \int v du$
$I = -(e^{x\cos x}) - \int (-\cos x).e^xdx$
$\;\;= - (e^{x\cos x}) +\int e^{x\cos x}dx$
Again using the same method of integration
Step 3:
Let $u = e^x ; du = e^xdx$
Let $dv = \cos xdx$; hence $v = \sin x$
-$(e^x\cos x) + e^{x\sin x} - \int e^{x\sin x} dx$
$I = e^{x\sin x} - e^{\cos x} -I$
$2I = e^{\large x\sin x} - e^{\large\cos x}$
$I =\large\frac{ [e^{\sin x }- e^{x\cos x}]}{2}$
$y =\large\frac{ [e^{x\sin x} - e^{x\cos x}]}{2 }$$+ C$
Step 4:
Now to evaluate the value of C let us take $x=0$ and $y =0$
$0 =\large\frac{[ e^0\sin 0 -e^0\cos 0]}{2}$$ +C$
$C=\large\frac{ 1}{2}$
Substituting the value of $C$we get,
$2y - 1 = e^{\large x(\sin x - \cos x)}$
This is the required solution.
answered Aug 14, 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
...