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Differential Equations

Question
CBSEENMA12035699
Wired Faculty App

Solve the differential equation left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent

Solution

Given differential equation is:
left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space equals space fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
This is a linear differential equation of the form
dy over dx plus Py space equals space straight Q
where space straight P space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space and space straight Q space equals fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction Therefore comma space straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent Thus space the space solution space is comma straight y left parenthesis straight I. straight F right parenthesis space equals space integral straight Q thin space left parenthesis straight I. straight F right parenthesis space dx rightwards double arrow straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals integral fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent dx Substitute space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space straight t semicolon
straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space cross times fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dt Thus comma space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space integral tdt rightwards double arrow space space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals straight t squared over 2 plus straight C rightwards double arrow space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses squared over 2 plus straight C 

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