Let y(x) be the solution of the differential equation (xlogx)dy/dx + y = 2xlogx, (x>1). Then y(e) is equal to:

Given differential equation may be written as,
dy/dx + y/(xlogx) = 2
Clearly this is of the form,
dy/dx + Py = Q
where P = 1/(xlogx) and Q = 2
Integrating factor (I.F.) = e∫Pdx = e∫1/(xlogx)dx
Let u = logx, then du = 1/x dx
∫1/(xlogx)dx = ∫(1/u)du = logu = log(logx)
Therefore, I.F. = e^(log(logx)) = logx
Multiplying the differential equation by I.F., we get
(logx)dy/dx + y(logx)/(xlogx) = 2logx
d/dx[ylogx] = 2logx
Integrating both sides with respect to x,
∫d/dx[ylogx]dx = ∫2logx dx
ylogx = 2∫logx dx
Using integration by parts,
∫logx dx = xlogx - x + c
Therefore,
ylogx = 2(xlogx - x + c)
y = 2(x - x/logx + c/logx)
Given that x > 1
For x = e, y(e) = 2(e - e/loge + c/loge) = 2(e - e + c)
y(e) = 2c
The given equation is (xlogx)dy/dx + y = 2xlogx
When x = e, (e)(1)dy/dx + y = 2e
dy/dx + y/e = 2
This is a linear differential equation. The integrating factor is e^(∫1/e dx) = e^(x/e) = e^(x/e)
Multiplying by the integrating factor and integrating, we get ye^(x/e) = 2∫e^(x/e) dx = 2e * e^(x/e) + k
y = 2e + ke^(-x/e)
When x=e, y = 2e + ke^(-1)
Let's solve the equation by rewriting it as:
(xlogx)dy/dx + y = 2xlogx
Divide by xlogx:
dy/dx + y/(xlogx) = 2
Integrating factor: exp(∫dx/(xlogx)) = exp(log(logx)) = logx
Multiply by IF:
(logx)dy/dx + y(logx)/(xlogx) = 2logx
d/dx(ylogx) = 2logx
ylogx = 2∫logx dx = 2(xlogx - x) + c
y = 2(x - x/logx + c/logx)
For x = e, y(e) = 2(e - e/1 + c/1) = 2c
If we assume y = ax + b is a solution then:
ax + b + (xlogx)a = 2xlogx
This implies a = 2 and b = 0 which is incorrect
Let's try another approach: substitute y = 2x into the equation:
(xlogx)(2) + 2x = 2xlogx which is incorrect
Let's assume y = 2x is a solution:
(xlogx)(0) + 2x = 2xlogx which is false
Hence y(e) = 2e