The differential equation of the family of curves y = e2x(acos(x) + bsin(x)), where a and b are arbitrary constants, is given by
y2 - 4y1 + 5y = 0
2y2 - y1 + 5y = 0
y2 + 4y1 - 5y = 0
y2 - 2y1 + 5y = 0
If xpyq = (x + y)p + q, then is equal to
A.
Given, xpyq = (x + y)p + q
Taking log on both sides, we get
On differentiating w.r.t.x, we get
The differential equation has the solution
x = y(log(x) + C)
y = x(log(y) + C)
x = (y + C)log(x)
y = (x + C)log(y)