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Question
Find the particular solution of the differential equation log(dy/dx)= 3x + 4y, given that y = 0 when x = 0.
Solution
Consider the differential equation,
`log(dy/dx)=3x+4y`
Taking exponent on both the sides, we have
`e^log(dy/dx)=e^(3x+4y)`
`=>dy/dx=e^(3x+4y)`
`=>dy/dx=e^(3x).e^(4y)`
`=>dy/(e^(4y))=e^(3x)dx`
Integration in both the sides, we have
`intdy/e^4y=inte^(3x)dx`
`e^(-4y)/(-4)=e^(3x)/3+C`
We need to find the particular solution.
We have, y=0, when x=0
`1/(-4)=1/3+C`
`=>C=-1/4-1/3`
`=>C=(-3-4)/12=-7/12`
Thus, the solution is `e^(3x)/3+e^(-4y)/4=7/12`
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