Find the particular solution of the differential equation `log ((dy)/(dx))=3x+4y` given that `y = 0` when `x = 0`.

Explanation

⇒`e^−4y dy=e^3x dx`

Integrating, we have

`e^−4y/-4 − e^3x/3+c=0`


`⇒3e^−4y+4e^3x−12c=0`

Since `x=0 when y=0`, substituting both the values, we have

`3+4−12c=0`

`⇒12c=7`

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