q = 2 z -cos z + c
when z = 0, q = 5
5 = -1+c
c = 6
so
q = 2z -cos z +6
solve the initial value problems:
1- dq/dz = 2 + sinz, q(0)=5
2-dy/dx = (x^2 -2)x, y(2)=4
3-dy/dx = 4e^5^x + x, y(0)=4
4- dy/dx 3x + 2, y(3)= 2
2 answers
dy/dx = (x^2 -2)x, y(2)=4
dy/dx = x^3 - 2x
y = (1/4)x^4 - x^2 + c
4 = (1/4)2^4 - 2^2 + c
that should get you started
dy/dx = x^3 - 2x
y = (1/4)x^4 - x^2 + c
4 = (1/4)2^4 - 2^2 + c
that should get you started