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Vector Algebra

Question
CBSEENMA12035684
Wired Faculty App

Find the particulars solution of the differential equation left parenthesis 1 plus straight x squared right parenthesis dy over dx equals open parentheses straight e to the power of straight m space tan to the power of negative 1 straight x end exponent end exponent minus straight y close parentheses comma space given space that space straight y space equals space 1 space when space straight x space equals space 0.

Solution
left parenthesis 1 plus straight x squared right parenthesis dy over dx space equals space straight e to the power of straight m space tan to the power of negative 1 straight x end exponent end exponent minus straight y rightwards double arrow dy over dx equals fraction numerator straight e to the power of mtan 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 minus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction equals fraction numerator straight e to the power of mtan 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 straight P space equals fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction comma space straight Q equals fraction numerator straight e to the power of mtan 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 straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space space space space space equals straight e to the power of integral fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction end exponent space space space space space space equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent
Thus the solution is
ye to the power of integral Pdx end exponent space equals integral Qe to the power of integral Pdx end exponent dx rightwards double arrow space space straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space space equals integral fraction numerator straight e to the power of straight m space 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 straight e to the power of tan to the power of negative 1 end exponent straight x end exponent. dx rightwards double arrow space straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space integral fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis 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 dx.... left parenthesis straight i right parenthesis integral fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis 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 dx space.... left parenthesis ii right parenthesis Let space left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x space equals straight z fraction numerator left parenthesis straight m plus 1 right parenthesis over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dz fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dz fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction equals space fraction numerator dz over denominator left parenthesis straight m plus 1 right parenthesis end fraction
Substituting in (ii),
  fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis end fraction integral straight e to the power of straight z dz equals fraction numerator straight e to the power of straight z over denominator left parenthesis straight m plus 1 right parenthesis end fraction equals fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction
Substituting in (i),
rightwards double arrow straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C space.... left parenthesis iii right parenthesis Putting space straight y space equals space 1 space and space straight x space equals space 1 comma space in space the space above space equation comma space space rightwards double arrow space space straight y space cross times space straight e to the power of tan to the power of negative 1 end exponent 1 end exponent space equals fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent 1 end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C space rightwards double arrow 1 cross times straight e to the power of straight pi over 4 end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis end exponent begin display style straight pi over 4 end style over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C therefore straight C space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis begin display style straight pi over 4 end style end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction minus straight e to the power of straight pi over 4 end exponent
Particular solution of the D.E. is straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis begin display style straight pi over 4 end style end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction minus straight e to the power of straight pi over 4 end exponent
 

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

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