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 general solution $\large\frac{dy}{dx}=$$\sin^{-1}x$

$\begin{array}{1 1}y = x\sin^{-1}x + \sqrt { 1 - x^2} + C \\ y = x\cos^{-1}x - \sqrt { 1 - x^2} + C \\y = x\tan^{-1}x + \sqrt { 1 + x^2} + C \\ y = x\cos^{-1}x + \sqrt { 1 + x^2} + C\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Integration by parts is $\int udv = uv - \int vdu$
Step 1:
Given:$\large\frac{ dy}{dx} $$= \sin^{-1}x$
Seperating the variables we get
$dy = \sin^{-1}x\; dx$
Integrating on both sides we get
$\int dy = \int\sin^{-1}xdx$
Integrating the RHS by parts using the information in the tool box we get,
Let $u = \sin^{-1}x\; du =\large\frac{ 1}{\sqrt{1-x^2}}$ and $dv = dx$ hence $x = v$
Step 2:
substituting in the formula we get
$\int y = x\sin^{-1}x -\int\large\frac{ xdx}{\sqrt{1-x^2 }}$
applying the substitution method we get,
Let $1-x^2 = t$, hence $-2xdx = dt$ or $xdx = \large\frac{-dt}{2}$
substituting this
$y = x\sin^{-1}x -\int \large\frac{(-1/2)dt}{\sqrt { t}}$
$y = x\sin^{-1}x +\large\frac{\Large\frac{ 1}{2 \sqrt t}}{\Large\frac{1}{2}}$
$y = x\sin^{-1}x + \sqrt { 1 - x^2} + C$
answered Aug 15, 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
...