a) move the x^4 over, take the square root of each side.
dy/dx=x^2
y= 1/3 x^3 +C or y=-1/3 x^3-C
Specific solution? It looks to me that both will work.
dy/dx=x^2 dy/dx=-x^2
y'^2=x^4 y'^2=x^4
and if you pursue it, there are two imaginary solutions also.
b)dy/y= dx*sinx
lny=- cosx
Find the specific solution for the following differential equations:
a) (dy/dx)^2 - x^4 = 0 given the initial value y(1) = 0
b) dy/dx = ysinx given the initial value y(pi) = 1
1 answer