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Differential Equations

Question
CBSEENMA12033100

Solve the following initial value problem:
(x + y + 1)2 dy = dx, y ( –1) = 0

Solution
The given differential equation is
                  (x + y + 1)dy = dx
or                  dy over dx space equals space fraction numerator 1 over denominator left parenthesis straight x plus straight y plus 1 right parenthesis squared end fraction
Put      straight x plus straight y plus 1 space equals space straight t comma space space space space therefore space space space space 1 plus dy over dx space equals dt over dx space space space or space space space dy over dx equals space dt over dx minus 1
therefore space space space space given space equation space becomes space dt over dx minus 1 space equals 1 over straight t squared
therefore space space space space space space space space space space space dt over dx equals 1 plus 1 over straight t squared
or               dt over dx space equals fraction numerator straight t squared plus 1 over denominator straight t squared end fraction
Separating the variables and integrating, we get,
                   integral fraction numerator straight t squared over denominator straight t squared plus 1 end fraction dt space equals space integral dx space space space space or space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus straight t squared end fraction close parentheses space dt space equals space integral 1. space dx
therefore space space space space space space straight t minus tan to the power of negative 1 end exponent straight t space equals space straight x plus straight c space space space space or space space space left parenthesis straight x plus straight y plus 1 right parenthesis space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight x plus straight c
therefore space space space space straight y plus 1 minus tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight c
Now          straight y left parenthesis negative 1 right parenthesis space equals space 0 space space space space space space space space rightwards double arrow space space space space straight y space equals space 0 space space space when space straight x space equals space minus 1
therefore space space space space 0 plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis negative 1 plus 0 plus 1 right parenthesis space equals space straight c space space space space space space rightwards double arrow space space space space 1 minus tan to the power of negative 1 end exponent 0 space equals space straight c
therefore space space space space space space space space space space space space space space 1 minus 0 space equals space straight c space space space space space space space space rightwards double arrow space space space space straight c space equals space 1
therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight y plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space 1
therefore space space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight y space space space space space or space space space straight x plus straight y plus 1 space equals space tan space straight y
which is required solution. 

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