For the equation dy + 7x2y = 0 , if y(0) = 3 / 7 ,dx then

For the equation dy + 7x²y = 0 , if y(0) = 3 / 7 ,
dx

then the value of y(1) is

From question, we have

	dy	+ 7x²y = 0 , y(0) =	3	, y(1) = ?
	dx		7

	dy	= -7x²y
	dx

	dy	= -7x²dx
	y

ln y = -	7x³	+ C
	3

y = e^{{ (-7x³ / 3) + C}}

y(0) = e^C =	3
	7

C = ln		3
		7

ln y = -	7x³	+ ln		3
	3			7

ln y - ln		3		= -	7x³
		7			3

ln		7y		= -	7x³
		3			3

y =	3	e ^{(-7x³/ 3)}
	7

∴ y(1) =	3	e ^{(-7/ 3)}
	7

3	e^{-7 / 3}
7

3	e^{-3 / 7}
7

7	e^{-3 / 7}
3

7	e^{-7 / 3}
3