Question

Solve the following differential equation:
$square root of 1 minus straight y squared end root space dx equals left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy$

Solution

The given differential equation is
   $square root of 1 minus straight y squared end root space dx space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy space space space or space space square root of 1 minus straight y squared end root space dy over dx space equals sin to the power of negative 1 end exponent straight y minus straight x$
or    $dx over dy space equals fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction minus fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction straight x space space space space or space space space dx over dy plus fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction straight x space equals space fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction$
Comparing it with $dy over dx plus Px space equals space straight Q comma$ we get,
   $straight P space equals space fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction comma space space straight Q space equals space fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction integral straight P space dy space equals space integral fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction dy space equals space sin to the power of negative 1 end exponent straight y comma space space space therefore space space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent$
Solution of differential equation is
   $straight x space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dy end exponent dy plus straight c$
or $straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space integral fraction numerator left parenthesis sin to the power of negative 1 end exponent space straight y right parenthesis over denominator square root of 1 minus straight y squared end root end fraction. space straight e to the power of sin to the power of negative 1 end exponent straight y space end exponent dy plus straight c$ ...(1)
Let $straight I space equals space integral fraction numerator open parentheses sin to the power of negative 1 end exponent space straight y close parentheses over denominator square root of 1 minus straight y squared end root end fraction. space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent dy$
Put $sin to the power of negative 1 end exponent straight y space equals straight t comma$    $therefore space space space space space fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction dy space equals space dt$
$therefore space space space space space space space space space space space space space space space space straight I space equals space integral straight t. space straight e to the power of straight t space dt space space space space space space space space space space space space space space space space space space space space space space equals space straight t space straight e to the power of straight t. space integral 1. space straight e to the power of straight t space dt space equals space straight t space straight e to the power of straight t space minus space straight e to the power of straight t space equals space left parenthesis straight t minus 1 right parenthesis space straight e to the power of straight t space space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus space 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent end exponent straight y space space space space space space space space space space space space space space space space space space space space space space$
$therefore space space from space left parenthesis 1 right parenthesis comma space space straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent plus straight c$ ...(2)
Now y(0) = 0 $rightwards double arrow space space space space space straight y space equals space 0 space space when space straight x space equals space 0$
$therefore$    $0 space equals space left parenthesis sin to the power of negative 1 end exponent 0 minus 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent 0 end exponent plus straight c$
$rightwards double arrow$    $0 space equals space left parenthesis 0 minus 1 right parenthesis space plus space straight c space space space space space space space space rightwards double arrow space space space straight c space equals space 1$
$therefore$ from (2), solution is
   $straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus space 1 right parenthesis space space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent plus 1$
or $straight x equals space sin to the power of negative 1 end exponent straight y space minus space 1 plus space straight e to the power of negative sin to the power of negative 1 end exponent straight y end exponent space space space space space space space space or space space space space straight x plus 1 minus space sin to the power of negative 1 end exponent straight y space equals space straight e to the power of negative sin to the power of negative 1 end exponent straight y end exponent$

For Daily Free Study Material Join wiredfaculty

Differential Equations

Solve the following differential equation:
$square root of 1 minus straight y squared end root space dx equals left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy$

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction