Eduprobe
Question:
If y(x) satisfies the differential equation y' - ytanx = 2xsecx and y(0) = 0, then:
y(π/4) = π/8√2
y'(π/4) = π/18
y(π/3) = π/9
y'(π/3) = 4π/3 + 2π/3√3
Solution:
dy/dx - ytanx = 2xsecx
cosx dy/dx + (-sinx)y = 2x
d/dx(ycosx) = 2x
y(x)cosx = x² + c, where c = 0
Since y(0) = 0.
When x = π/4, y(π/4) = π/8√2;
When x = π/3, y(π/3) = 2π/9;
When x = π/4, y'(π/4) = π/8√2 + π√2
When x = π/3, y'(π/3) = 2π/3√3 + 4π/3.