Differential Equations

Solve the following differential equation: (1 + y 2 )dx = (tan – 1 y – x) dy - Wired Faculty

Question

Solve the following differential equation:
(1 + y²)dx = (tan^{– 1} y – x) dy

Solution

The given equation is (1 + y²)dx = (tan^{– 1} y – x) dy
or $left parenthesis 1 plus straight y squared right parenthesis space dx over dy space equals space tan to the power of negative 1 end exponent straight y minus straight x space space space or space space left parenthesis 1 plus straight y squared right parenthesis space dy over dx plus straight x space equals space tan to the power of negative 1 end exponent straight y$
or $dx over dy plus fraction numerator 1 over denominator 1 plus straight y squared end fraction straight x space equals space fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction$
$therefore$ Comparing it with $dy over dx plus Px space equals space straight Q$ , we get,
$straight P space equals fraction numerator 1 over denominator 1 plus straight y squared end fraction comma space space straight Q space equals fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction comma space space space integral straight P space dy space equals space integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space tan to the power of negative 1 end exponent straight y$
$therefore space space space straight I. straight F. space equals space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent therefore space space space solution space of space equation space is$
$straight x. space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space space integral fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction straight e to the power of tan to the power of negative 1 end exponent straight y end exponent dy plus straight c space space space space space space space space space open square brackets because space space space straight x space straight e to the power of integral straight P space dy end exponent space equals space integral straight Q space straight e to the power of integral straight P space dy end exponent dy space plus straight c close square brackets$
Put $tan to the power of negative 1 end exponent straight y space equals straight t space space space space space space space space space space space space space space space space space space space therefore space space space fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space dt$
$therefore space space straight x space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space integral straight t space straight e to the power of straight t space dt space plus space straight c space equals space ke to the power of straight t minus space integral straight e to the power of straight t. space dt space plus space straight c space equals space straight t space straight e to the power of straight t space minus space straight e to the power of straight t plus straight c therefore space space space straight x space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent left parenthesis tan to the power of negative 1 end exponent straight y space space minus 1 right parenthesis space plus straight c$

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