Question

Find the particular solution of the differential equation $\log (\frac{dy}{dx})=3x+4y$, given that y = 0 when x = 0.

Answer 1

3×0+4×0
0+0
=0
it is the correct answer

Answer 2

Answer:

particular solution:at x=0,y=0

$log(\frac{dy}{dx} )=3x+4y\\\\is\\\\\frac{e^{-4y}}{-4} =\frac{e^{3x}}{3}-\frac{7}{12}$

Step-by-step explanation:

To find the solution of the given differential equation,we can solve this by variable separation method

taking antilog both sides

$log(\frac{dy}{dx} )=3x+4y\\ \\ \\ \frac{dy}{dx} =e^{(3x+4y)} \\ \\ \\ \frac{dy}{dx} =e^{3x}.e^{4y}\\ \\ \\ \frac{dy}{e^{4y}}=e^{3x}\:dx\\ \\ \\ e^{-4y}dy=e^{3x}\:dx$

now integrate both sides

$\int\:e^{-4y}dy=\int\:e^{3x}\:dx\\ \\ \frac{e^{-4y}}{-4} =\frac{e^{3x}}{3}+C$

is the general solution   of the given differential equation,so for particular solution

put x=y=0

$\frac{e^{-4.0}}{-4} =\frac{e^{3.0}}{3}+C\\ \\ \frac{-1}{4} -\frac{1}{3}=C\\\\C=\frac{-7}{12}$

particular solution:at x=0,y=0

$log(\frac{dy}{dx} )=3x+4y\\\\is\\\\\frac{e^{-4y}}{-4} =\frac{e^{3x}}{3}-\frac{7}{12}$