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Vector Algebra

Question
CBSEENMA12033075

Find the particular solution of the differential equation log space open parentheses dy over dx close parentheses space equals space 3 straight x plus 4 straight y given that y = 0 when x = 0.

Solution

The given differential equation is
log open parentheses dy over dx close parentheses space equals space 3 straight x plus 4 straight y space space space space or space space space space space dy over dx equals space straight e to the power of 3 straight x plus 4 straight y end exponent space space space or space space space space space dy over dx space equals straight e to the power of 2 straight x end exponent. space straight e to the power of 4 straight y end exponent
Separating the variables, dy over straight e to the power of 4 straight y end exponent space equals straight e to the power of 3 straight x end exponent space dx space or space space space integral straight e to the power of negative 4 straight y end exponent dy space equals space integral straight e to the power of 3 straight x end exponent space dx
therefore space space space fraction numerator straight e to the power of negative 4 straight y end exponent over denominator negative 4 end fraction space equals space straight e to the power of 3 straight x end exponent over 3 plus straight c                                    ...(1)
when y = 0,  x = 0,   then from (1), we get,
                negative straight e to the power of 0 over 4 space equals space straight e to the power of 0 over 3 plus straight c space space space space or space space space minus 1 fourth space equals space 1 third plus straight c
therefore space space space space space space straight c space equals space minus 1 fourth minus 1 third space equals space minus 7 over 12
therefore space space from (1),  1 fourth straight e to the power of negative 4 straight y end exponent space equals space 1 third straight e to the power of 3 straight x end exponent space minus space 7 over 12 which  is required solution.