Asked by Suzie
Find the solution to the differential equation dy/dx = (x^3)/(y^2),
where y(2) = 3
I have:
The integral of: dy(y^2) = the integral of: dx(x^3)
(y^3)/3 = (x^4)/4
The answer is supposed to come out to be y= the cube root of: ( 3/4x + 15 )
Thanks
The general solution is
(y^3)/3 = (x^4)/4 + C
where C is any constant. If y = 3 when x=2, then
27/3 = 16/4 + C
C = 9 - 4 = 5
y^3 = (3/4) x^4 + 3C = (3/4)x^4 + 15
y = cube root of [(3/4)x^4 + 15]
I believe you left out an exponent ^4 in the answer.
where y(2) = 3
I have:
The integral of: dy(y^2) = the integral of: dx(x^3)
(y^3)/3 = (x^4)/4
The answer is supposed to come out to be y= the cube root of: ( 3/4x + 15 )
Thanks
The general solution is
(y^3)/3 = (x^4)/4 + C
where C is any constant. If y = 3 when x=2, then
27/3 = 16/4 + C
C = 9 - 4 = 5
y^3 = (3/4) x^4 + 3C = (3/4)x^4 + 15
y = cube root of [(3/4)x^4 + 15]
I believe you left out an exponent ^4 in the answer.